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Problem. Let $a, b, c \in \mathbb{R}$. Prove that $$ \sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geqslant\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}. $$

I think that this inequality is strong, though I do not have knowledge of many techniques. There goes my work:
Positive variables only make the inequality stronger, hence suppose $a,b,c\geqslant0$ $$ \sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geqslant\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)} $$By squaring, $$ \Rightarrow \sqrt{(a^2+5b^2)(b^2+5c^2)}+\sqrt{(b^2+5c^2)(c^2+5a^2)}+\sqrt{(c^2+5a^2)(a^2+5b^2)}\geq2(a+b+c)^2 $$The $LHS$ $$= \sqrt{\sum_{cyc}{5b^4 + 31a^2b^2 + 2\left(a^2 + 5b^2\right) \left(\sqrt{\left(b^2 + 5c^2\right) \left(c^2 + 5a^2\right)}\right)}} $$$$ \geqslant \sqrt{\sum_{cyc}{5b^4 + 31a^2b^2 + 2(a^2 + 5b^2)(bc + 5ca)}} $$ Now we are only left to prove that $$ \sum_{cyc}{5b^4 + 31a^2b^2 + 52a^2bc + 10a^3c + 10a^3c} \geqslant \sum_{cyc}{4a^4 + 16(a^3b + ab^3) + 24a^2b^2 + 48a^2bc} $$$$ \sum_{cyc}{a^4 + 7a^2b^2 + 4a^2bc - 6(a^3b + ab^3)} \geqslant 0 $$ The last inequality is wrong for $(a,b,c) = (1,1,0)$. Cauchy Schwarz looks fine but I am not able to find a way.
I found this inequality posted by arqady on aops forum.
Please help!

River Li
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Book Of Flames
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    Say me please, who is the author of this inequality? – Michael Rozenberg Oct 04 '20 at 05:03
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    AM-GM in 2nd step for LHS gives the result $6(a^2+b^2+c^2) \ge 2(a+b+c)^2$, which is easily provable to be true. – Aniruddha Deb Oct 04 '20 at 05:04
  • @MichaelRozenberg You have founded it. – Book Of Flames Oct 04 '20 at 05:06
  • And @AniruddhaDeb You are loosening the inequality, by applying AM-GM there. So it is not proof. – Book Of Flames Oct 04 '20 at 05:09
  • @MichaelRozenberg where is your inequality written? I bought the first Cirtoaje book, you have some entries there. – Will Jagy Oct 04 '20 at 13:13
  • @WillJagy This inequality is from one of his posts on AoPS site. – Book Of Flames Oct 04 '20 at 13:28
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    And @Michael Rozenberg, I apologize if I have done something bad to you, but don't you have a proof of this inequality? – Book Of Flames Oct 04 '20 at 13:28
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    @Will Jagy I created it yesterday. See here: https://artofproblemsolving.com/community/c6h2290905p18034221 – Michael Rozenberg Oct 04 '20 at 13:29
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    @MichaelRozenberg In that case I would say it should not be posted here so soon. it is rude to do so. The foreword to the book indicates AoPS as a primary source, with Vasc as Cirtoaje's username. I am not positive, it would appear your user name there is arqady or possibly Nguyenhuyen. The first one, you did put your name in the little profile page. – Will Jagy Oct 04 '20 at 13:36
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    @Will Jagy My username in AoPS it's arqady. I don't see any problem to post this inequality here. By the way, in this site there are very many my inequalities without any link that it's mine.. What can I do with it? This is our life, apparently... :) – Michael Rozenberg Oct 04 '20 at 13:59
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    @BookOfFlames well this is a very strict inequality. And by the way, your last "wrong" inequality could be written as $(a^2+b^2+c^2-ab-bc-ca)(a^2+b^2+c^2-5(ab+bc+ca)) \geq 0$. I don't know if that can help us see anything? – Đào Minh Dũng Oct 06 '20 at 05:24
  • @ĐàoMinhDũng That leaves us with $a^2 + b^2 +c^2 \geqslant 5(ab + bc + ca)$. It is wrong, but still, I think that following my approach and squaring again will give more strict result. But I have another estimation for this inequality: $\sum_{cyc}{(a^2 + 5b^2)\sqrt{\left(b^2 + 5c^2\right)\left(c^2 + 5a^2\right)}}\geqslant\frac{ \left(\sum_{cyc}{\sqrt{a^2 + 5b^2}}\right)^2}{\sum_{cyc}{\frac{1}{a^2 + 5b^2}}}$, still I am not sure it will help. – Book Of Flames Oct 06 '20 at 05:56
  • Use rational bounds for $\sqrt{a^2+5b^2}$, then Buffalo Way works. But it is a computer solution (very very complicated by hand). – River Li Oct 06 '20 at 16:02
  • @RiverLi What do you mean by rational bounds? – Book Of Flames Oct 06 '20 at 16:07
  • @River Li I also got that it's enough to prove an inequality with one radical and little coefficients, but by another idea. In the rest I still don't see a human proof. – Michael Rozenberg Oct 06 '20 at 21:19
  • @BookOfFlames Please see my previous answers: https://math.stackexchange.com/questions/1997333/prove-that-sum-limits-cyc-sqrt3a24bc-geq-sqrt345abacbc/3303551#3303551, https://math.stackexchange.com/questions/2036890/prove-that-sum-limits-cyc-sqrta210bc-geq-frac12-sum-limits-cyc-sqr/3372011#3372011 – River Li Oct 07 '20 at 01:43
  • @MichaelRozenberg A fun thing I noticed. Iff $a,b,c \in \mathbb{R}$ and If instead of 8, it was k while $$k \in [-8,8]$$ the inequality holds true, anything other than that (even if k is < -8), the inequality is false for some a,b,c – Anindya Prithvi Oct 07 '20 at 07:33
  • @AnindyaPrithvi That means the inequality is very strict; It will not be obliterated by usual means. – Book Of Flames Oct 07 '20 at 07:35
  • @MichaelRozenberg Often, if we eliminate the square root signs in some inequalities, the resulting polynomial inequalities are complicated and hard to prove by hand. – River Li Oct 07 '20 at 15:39
  • I think one can WLOG consider the case $a^2 + b^2 + c^2 = 1$, due to homogeneity. I ran simulations on a very dense unit sphere in MATLAB and it appears to hold, but this of course is a dirty way to prove it. – V.S.e.H. Oct 08 '20 at 01:34
  • @MichaelRozenberg Also, for a weak one, $\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\ge 3\sqrt{a^2+b^2+c^2 + ab + bc + ca}$, there is a nice solution by hand. – River Li Oct 08 '20 at 02:09
  • @River Li $\sum\limits_{cyc}\sqrt{a^2+5b^2}\geq\sqrt{\frac{1}{3}\sum\limits_{cyc}(28a^2+26ab)}$ also has a nice proof, but it's stronger. See my post. – Michael Rozenberg Oct 08 '20 at 03:07
  • @MichaelRozenberg Nice. I used something similar, that is $a^2 + 5b^2 = \frac{(a+5b)^2}{6} + \frac{5}{6}(a-b)^2$; and C-S to get $\sqrt{a^2+5b^2}\sqrt{b^2+5c^2} \ge \frac{1}{6}(a+5b)(b+5c) + \frac{5}{6}(b-a)(b-c)$. – River Li Oct 08 '20 at 05:13
  • @River Li Yes, it's nice and it helps to remove three radicals in the starting inequality. We need now to remove the last radical. I got that $\sqrt{\frac{\prod\limits_{cyc}(a^2+5b^2)}{6}}\geq\frac{\sum\limits_{cyc}(a^3b+5b^3c+15a^2b^2+15a^2bc)}{6(a+b+c)}.$ and it's enough to prove that: $\sum\limits_{cyc}(5a^4+31a^2b^2)+\frac{2\sum\limits_{cyc}(a^3b+5a^3c+15a^2b^2+15a^2bc)\sqrt{\sum\limits_{cyc}(14a^2+13ab)}}{3(a+b+c)}\geq4(a+b+c)^4.$ I can not check, if it's true. But if it's true, I can prove it. – Michael Rozenberg Oct 08 '20 at 05:22
  • @MichaelRozenberg Please check $\sqrt{\frac{\prod\limits_{cyc}(a^2+5b^2)}{6}}\geq\frac{\sum\limits_{cyc}(a^3b+5b^3c+15a^2b^2+15a^2bc)}{6(a+b+c)}$ for $a=2, b=1, c=0$? – River Li Oct 08 '20 at 07:18
  • @River Li I made a typo. Change $b^3c$ on $a^3c$ as in the second inequality. If the second inequality is true, it would be beautiful! Thank you for your interest. – Michael Rozenberg Oct 08 '20 at 07:23
  • @MichaelRozenberg Please check the second inequality for $c = 0, b=1, a=6$? – River Li Oct 08 '20 at 10:10
  • Yes I see. Thank you! Now I see that I could find it by myself. Id est, we still have no a nice proof. :( – Michael Rozenberg Oct 08 '20 at 10:20
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    @MichaelRozenberg You may also notice the identity $a^2 + kb^2 = \frac{(a+kb)^2}{k+1} + \frac{k}{k+1}(a-b)^2$. Also we have $\sqrt{a^2 + kb^2} = \frac{a+kb}{\sqrt{k+1}}
    • \frac{\frac{k}{k+1}(a-b)^2}{\sqrt{a^2 + kb^2} + \frac{a+kb}{\sqrt{k+1}}}$ from which we can obtain some bounds. I used it for $k=3$ inequality.
     – River Li Oct 08 '20 at 15:51
  • @River Li $k=5$, you mean. Yes, of course. We can get it by C-S. See my post. But it does not so help. The second is interesting. – Michael Rozenberg Oct 08 '20 at 15:53
  • When $k=3$, it is $\sqrt{a^2 + 3b^2} = \frac{a+3b}{2} + \frac{3(a-b)^2/4}{\sqrt{a^2+3b^2} + (a+3b)/2}$. For both $k=3, 5$, by good bounds, there is a computer solution. I can not prove it by hand. – River Li Oct 08 '20 at 15:56
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    I have found the effective approach, but too late. – Yuri Negometyanov Oct 13 '20 at 03:18
  • @YuriNegometyanov you still have 27 mins – Hari Ramakrishnan Sudhakar Oct 13 '20 at 03:25
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    @YuriNegometyanov Maybe I'm not able to provide you with the bounty amount but you can post your answer at least. – Book Of Flames Oct 13 '20 at 16:29
  • The best coefficient is about $\sqrt{10.14 \sum a^2 + 7.86 \sum ab}$, with trouble near $(a,b,c) = (0.30, 2.07, 0.63)$ – Chen Dan Oct 25 '25 at 02:28

3 Answers3

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Since $x\leq|x|$, it's enough to prove our inequality for non-negative variables.

Now, after squaring of the both sides we need to prove that $$\sum_{cyc}\sqrt{(a^2+5b^2)(b^2+5c^2)}\geq2(a+b+c)^2.$$ Also, $$\sum_{cyc}\sqrt{(a^2+5b^2)(b^2+5c^2)}=$$ $$=\sqrt{\sum_{cyc}\left((a^2+5b^2)(b^2+5c^2)+2(c^2+5a^2)\sqrt{(a^2+5b^2)(b^2+5c^2)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(5a^4+31a^2b^2+2\sqrt{\prod_{cyc}(a^2+5b^2)}\sqrt{a^2+5b^2}\right)}=$$ $$=\sqrt{\sum_{cyc}(5a^4+31a^2b^2)+2\sqrt{\prod_{cyc}(a^2+5b^2)}\sqrt{\sum_{cyc}\left(6a^2+2\sqrt{(a^2+5b^2)(b^2+5c^2)}\right)}}.$$ But by C-S $$2\sum_{cyc}\sqrt{(a^2+5b^2)(b^2+5c^2)}=$$ $$=\frac{1}{3}\sum_{cyc}\sqrt{((a+5b)^2+5(a-b)^2)((b+5c)^2+5(b-c)^2)}\geq$$ $$\geq\frac{1}{3}\sum_{cyc}((a+5b)(b+5c)+5(b-a)(b-c)).$$ Id est, it's enough to prove that: $$\sum_{cyc}(5a^4+31a^2b^2)+2\sqrt{\frac{1}{3}\prod_{cyc}(a^2+5b^2)\sum_{cyc}(28a^2+26ab)}\geq4(a+b+c)^4$$ or $$2\sqrt{\frac{1}{3}\prod_{cyc}(a^2+5b^2)\sum_{cyc}(28a^2+26ab)}\geq4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2),$$ which is obvious for $$4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)<0.$$ Thus, it's enough to prove our inequality for $$4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)\geq0.$$ Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Thus, we need to prove that: $$8\prod_{cyc}(a^2+5b^2)\sum_{cyc}(14a^2+13ab)\geq3\left(4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)\right)^2,$$ for which it's enough to prove that: $$8\prod_{cyc}(a^2+5b^2)\sum_{cyc}(14a^2+13ab)-3\left(4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)\right)^2\geq$$ $$\geq4\left(u^2-5uv+v^2\right)^2\left(4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)\right ).$$ Now, $$8\prod_{cyc}(a^2+5b^2)\sum_{cyc}(14a^2+13ab)-3\left(4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)\right)^2=$$ $$=19008(u^2-uv+v^2)a^6+5184(4u^3-5u^2v+15uv^2+4v^3)a^5+$$ $$+144(109u^4-458u^3v+867u^2v^2+742uv^3+109v^4)a^4+$$ $$+64(114u^5-677u^4v+529u^3v^2+2399u^2v^3+1088uv^4+114v^5)a^3+$$ $$+4(438u^6-2978u^5v+107u^4v^2+13656u^3v^3+20467u^2v^4+5822uv^5+438v^6)a^2+$$ $$+4(42u^7-269u^6v-191u^5v^2+2403u^4v^3+5523u^3v^4+5249u^2v^5+851uv^6+42v^7)a-$$ $$-3u^8+96u^7v-250u^6v^2+1288u^5v^3+1671u^4v^4+3368u^3v^5+1990u^2v^6+96uv^7-3v^8.$$ Also, $$4(a+b+c)^4-\sum_{cyc}(5a^4+31a^2b^2)=$$ $$=216a^4+288(u+v)a^3+4(31u^2+77uv+31v^2)a^2+$$ $$+(28u^3+82u^2v+82uv^2+28v^3)a-u^4+16u^3v-7u^2v^2+16uv^3-v^4.$$ Easy to see that $$19008(u^2-uv+v^2)\geq19008uv,$$ $$5184(4u^3-5u^2v+15uv^2+4v^3)\geq80899\sqrt{u^3v^3},$$ $$144(109u^4-458u^3v+867u^2v^2+742uv^3+109v^4)-$$ $$-4\left(u^2-5uv+v^2\right)^2\cdot216\geq99373u^2v^2,$$ $$64(114u^5-677u^4v+529u^3v^2+2399u^2v^3+1088uv^4+114v^5)-$$ $$-4\left(u^2-5uv+v^2\right)^2\cdot288(u+v)\geq35586\sqrt{u^5v^5},$$ $$4(438u^6-2978u^5v+107u^4v^2+13656u^3v^3+20467u^2v^4+5822uv^5+438v^6)-$$ $$-4\left(u^2-5uv+v^2\right)^2\cdot4(31u^2+77uv+31v^2)\geq-6165u^3v^3,$$ $$4(42u^7-269u^6v-191u^5v^2+2403u^4v^3+5523u^3v^4+5249u^2v^5+851uv^6+42v^7)-$$ $$-4\left(u^2-5uv+v^2\right)^2\cdot(28u^3+82u^2v+82uv^2+28v^3)\geq11491\sqrt{u^7v^7}$$ and $$-3u^8+96u^7v-250u^6v^2+1288u^5v^3+1671u^4v^4+3368u^3v^5+1990u^2v^6+96uv^7-3v^8-$$ $$-4\left(u^2-5uv+v^2\right)^2(-u^4+16u^3v-7u^2v^2+16uv^3-v^4)\geq5432u^4v^4.$$ Now, let $a=\sqrt{uv}t.$

Thus, it's enough to prove that: $$19008t^6+80899t^5+99373t^4+35586t^3-6165t^2+11491t+5432\geq0,$$ which is obvious.

Michael Rozenberg
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Probably not the proof you are looking for, but a proof nonetheless.

The inequality is really sharp, and I don't think that a manual solution exists. Concretely, I don't think that one can find a lower bound on the LHS, such that we can algebraically confirm that it upper bounds the RHS. However, it is easy to numerically verify that the inequality holds, and I hope that you can find this convincing.

Specifically, divide both sides by $\sqrt{a^2 + b^2 + c^2}$, then we're left with the equivalent inequality: $$ \sqrt{x^2 + 5y^2} + \sqrt{y^2 + 5z^2} + \sqrt{z^2 + 5x^2} \geq \sqrt{10 + 8(xy + yz + xz)}, $$ where $x = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, y = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, z = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$, and $x^2 + y^2 + z^2 = 1$. Furthermore, it has been established that we can safely assume that $x,y,z\geq 0$, so it is sufficient to verify the inequality on the surface $\{(x,y,z) \in\mathbb{R}^3 ~\vert~ x^2 + y^2 + z^2 = 1, x,y,z\geq 0\}$, which can be parameterized with $$x = \sin\theta\sin\omega,\quad y = \sin\theta\cos\omega,\quad z=\cos\theta,$$ with $(\theta,\omega)\in[0,\pi/2]\times[0,\pi/2]$.

Now, if one minimizes the function $$ h(\theta,\omega) = \sqrt{x^2 + 5y^2} + \sqrt{y^2 + 5z^2} + \sqrt{z^2 + 5x^2} - \sqrt{10 + 8(xy + yz + xz)}, $$ over the square $[0,\pi/2]\times[0,\pi/2]$, one then finds that it has a unique global minimum 0 at $x=y=z=\frac{1}{\sqrt{3}}$, or at $\theta \approx 0.9554,~ \omega = \pi/4$, see the figure below which shows the level sets of $h$.

Level sets of <span class=$h(\theta,\omega)$" />

This implies by homogeneity that the original inequality is equality only at $a=b=c$, and a strict inequality at all other values.

V.S.e.H.
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    How do you prove the unique minimization of $h(\theta,\omega)$ at the point you claim? The proof is not complete. – Hans Feb 20 '21 at 22:57
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Here is a sketch of proof :

We show the hardest case when $a\geq b\geq c $ and $5a^2+c^2\geq 5b^2+a^2\geq 5c^2+b^2$

If we show the following statement :

Let $a\geq b\geq c>0 $ and $5a^2+c^2\geq 5b^2+a^2\geq 5c^2+b^2$ and $n\geq 400$ a natural number then we have :

$$\frac{1}{n+1}(5a^2+c^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\geq \frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\quad(1)$$

And :

$$\Big(\frac{1}{n+1}(5a^2+c^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\Big)\Big(\frac{1}{n+1}(5b^2+a^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\Big)\geq \Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)^2\quad(2)$$

And :

$$\Big(\frac{1}{n+1}(5c^2+b^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\Big)\Big(\frac{1}{n+1}(5a^2+c^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\Big)\Big(\frac{1}{n+1}(5b^2+a^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\Big)\geq \Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)^3\quad(3)$$

And then applying Karamata's inequality we show :

$$ \sqrt{\frac{1}{n+1}(5a^2+c^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big)\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}}+\sqrt{\frac{1}{n+1}(5b^2+a^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big) \frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}}+\sqrt{\frac{1}{n+1}(5c^2+b^2)+\frac{n}{n+1}\Big(\frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}\Big) \frac{10(a^2+b^2+c^2)-8(ab+bc+ca)}{9}}\geqslant\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)} \quad(4)$$

Remains to apply this kind of inequality :

$$\sqrt{x}\frac{1}{k+1}+\sqrt{y}\frac{k}{k+1}\geq \sqrt{x\frac{1}{n+1}+y\frac{n}{n+1}}\quad (5)$$

Where $x,y>0$ and $n,k>0$ naturals numbers .

Applying $(4)$ to $(5)$ we get the desired inequality .

$(1)$ is trivial we prove $(2)$ now :

Due to homogeneity we put $a=1+p+q$ , $b=1+p$, $c=1$ and now we use WA .

See here for the LHS and here for the RHS of $(2)$ . If we substract each coefficient the remainders is positive wich prove the inequality $(2)$ .We can show $(3)$ by a similar way .

Hope you learn something from me and it helps you .

Regards Max

Barackouda
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