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Problem. Let $a,b,c\ge 0: a+b+c>0$ then prove that $$\sqrt{25a^2+144bc}+\sqrt{25b^2+144ca}+\sqrt{25c^2+144ab} \ge 5(a+b+c)+\frac{24(ab+bc+ca)}{a+b+c}.$$ Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim(t,t,0)$ or $(a,b,c)\sim(0,0,t)$ where $t>0.$

This inequality was unsolved on AoPS.

I've tried to use Holder inequality to eliminate radical sign without success. Also, there was an idea in the given link but it seems wrong.

Could you please give a hint solve the inequality? Thank you for your interest.

Michael Rozenberg
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    Notice that the right side has an upper bound of $13(a+b+c)$ (see https://math.stackexchange.com/questions/1044619/proving-abc2-3bccaab). I can't go any further for right now. – Debalanced Aug 12 '25 at 01:35
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    I have the feeling that there is a geometric interpretation for both sides, by letting $a=r\cot\frac{A}{2}$ and so on. Not sure though, let us wait for Micheal. – Jack D'Aurizio Aug 12 '25 at 01:46

4 Answers4

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Without loss of generality, assume $a\le b\le c$.

First we show that $$\sqrt{25b^2+144ca}+\sqrt{25c^2+144ab}\ge\sqrt{25(b+c)^2+288a(b+c)}.$$ By squaring both sides, this simplifies to $$\sqrt{25b^2+144ca}\cdot\sqrt{25c^2+144ab}\ge72ab+72ac+25bc,$$ for which $$\text{LHS}^2-\text{RHS}^2=144a(b-c)^2(25b+25c-36a)\ge0.$$ In the original problem, it remains to show that $$\sqrt{25a^2+144bc}+\sqrt{25(b+c)^2+288a(b+c)}\ge5\sum_{\rm cyc}a+\frac{24\sum ab}{a+b+c}.$$ It suffices to prove that $$25(b+c)^2+288a(b+c)\ge\left(5\sum_{\rm cyc}a+\frac{24\sum ab}{a+b+c}-\sqrt{25a^2+144bc}\right)^2,$$ or \begin{align*} \sqrt{25 a^2+144 b c} \left(\begin{array}c5 a^3+39 a^2 b+39 a^2 c+39 a b^2+102 a b c\\+39 a c^2+5 b^3+39 b^2 c+39 b c^2+5 c^3\end{array}\right)\ge\\ \left(\begin{array}c25 a^4+51 a^3 b+51 a^3 c+315 a^2 b^2+822 a^2 b c\\+315 a^2 c^2+a b^3+963 a b^2 c+963 a b c^2\\+a c^3+192 b^3 c+672 b^2 c^2+192 b c^3\end{array}\right). \end{align*} Squaring both sides leads to $48T\ge0$, where $$T=(b-a) (c-a) \left(\begin{array}c150 a^5 b+150 a^5 c+763 a^4 c^2\\+1088 a^2 b^4+\frac{48438}{5} a^2 b c^3+1162 a b^5\\+\frac{8116}{5} a b^4 c+\frac{1592}{5} a b c^4+655 a c^5\end{array}\right)+(c-a)^2 \left(\begin{array}c1526 a^4 b c+9011 a^3 b c^2+871 a^3 c^3\\+832 a^2 b^4+\frac{21747}{5} a^2 b c^3+668 a^2 c^4\end{array}\right)+(b-a)^2 \left(\begin{array}c763 a^4 b^2+2466 a^3 b^3+2700 a^3 b^2 c\\+1175 a^2 b^4+3704 a^2 b^3 c+\frac{32247}{5} a b c^4\end{array}\right)+(b-c)^2 \left(\begin{array}c832 a^5 c+323 a^4 b c+\frac{20734}{5} a^3 b^2 c+\frac{4588}{5} a^3 b c^2+\frac{6846}{5} a^2 b^3 c\\+\frac{19663}{5} a^2 b^2 c^2+\frac{14154}{5} a b^3 c^2+\frac{25748}{5} a b^2 c^3+507 a b c^4\\+75 b^5 c+552 b^4 c^2+1386 b^3 c^3+552 b^2 c^4+75 b c^5\end{array}\right)\ge0.$$

youthdoo
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Proof.

The inequality is obvious for $ab+bc+ca=0.$

For $ab+bc+ca>0$ the inequality is equivalent to$$\frac{bc}{\sqrt{25a^2+144bc}+5a}+\frac{ca}{\sqrt{25b^2+144ca}+5b}+\frac{ab}{\sqrt{25c^2+144ab}+5c} \ge \frac{ab+bc+ca}{6(a+b+c)}.$$ $*$ If $abc=0$ we can easily use AM-GM inequality.

$*$ If $abc>0$ we may use Cauchy-Schwarz inequality as $$\sum_{\mathrm{cyc}}\frac{bc}{\sqrt{25a^2+144bc}+5a}\ge \frac{(ab+bc+ca)^2}{\sum\limits_{\mathrm{cyc}}\left(bc\sqrt{25a^2+144bc}+5abc\right)}.$$ It's enough to prove$$6(a+b+c)(ab+bc+ca)\ge 15abc+bc\sqrt{25a^2+144bc}+ca\sqrt{25b^2+144ca}+ab\sqrt{25c^2+144ab}.$$ Also by C-B-S inequality, we'll prove $$6(a+b+c)(ab+bc+ca)-15abc\ge\sqrt{\left(ab+bc+ca\right)\left[25abc\left(a+b+c\right)+144\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\right)\right]}.\ \ \quad(*)$$ By squaring both side, we get SOS expression$$\sum a b \left(a - b\right)^{2} \cdot \left(36 a b + 11 a c + 11 b c\right) + 50\prod a \sum a \left(a - b\right) \left(a - c\right)\ge 0.$$ We can prove $(*)$ by $pqr$ method.

Let $a+b+c=p,\ ab+bc+ca=q=1,\ abc=r.$ The inequality $(*)$ becomes$$6p-15r\ge \sqrt{25pr+144(1-2pr)}$$or$$36p^2+83pr+225r^2-144\ge 0.$$ We just need to check the case $\sqrt{3}\le p<2.$

By using Schur of third degree inequality $r\ge \dfrac{4p-p^3}{9}>0$ we obtain \begin{align*} 36p^2+83pr+225r^2-144&\ge 36p^2+83p\cdot\frac{4p-p^3}{9}+225\cdot \left(\frac{4p-p^3}{9}\right)^2-144\\&=(p^2 - 4) (p^2 - 3) (25 p^2 - 108)\ge 0,\ \ \forall \sqrt{3}\le p<2. \end{align*} The inequality is proven. Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim(t,t,0)$ or $(a,b,c)\sim(0,0,t)$ where $t>0.$

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Since you only asked for a hint. Apply Jensen's Inequality to symmetrizes the problem.

coffee addict
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Supplement to @youthdoo's answer.

WLOG, assume that $a \le b \le c$. If $a = 0$, the desired inequality is true (easy). In what follows, assume that $a > 0$. WLOG, assume that $a = 1$.

It suffices to prove that, for all $b, c\ge 1$, $$\sqrt{25+144bc}+\sqrt{25(b+c)^2+288(b+c)} \ge 5(1 + b + c) + \frac{24(b + c+bc)}{1+b+c}.\tag{1}$$

We have \begin{align*} \sqrt{25+144bc} - 5 - \frac{24bc}{1+b+c} &= \frac{144bc}{\sqrt{25+144bc} + 5} - \frac{24bc}{1+b+c}\\ &= 2bc \left(\frac{72}{\sqrt{25+144bc} + 5} - \frac{12}{1+b+c}\right)\\ &\ge 2bc \left(\frac{72}{\sqrt{25+36(b+c)^2} + 5} - \frac{12}{1+b+c}\right)\\ &\ge (b + c)\left(\frac{72}{\sqrt{25+36(b+c)^2} + 5} - \frac{12}{1+b+c}\right) \tag{2} \end{align*} where we use $bc \le (b + c)^2/4$ and $bc + bc \ge b + c$ and $\frac{72}{\sqrt{25+36(b+c)^2} + 5} - \frac{12}{1+b+c}\ge 0$.

From (1) and (2), it suffices to prove that $$(b + c)\left(\frac{72}{\sqrt{25+36(b+c)^2} + 5} - \frac{12}{1+b+c}\right) + \sqrt{25(b+c)^2+288(b+c)}$$ $$ \ge 5(b + c) + \frac{24(b + c)}{1+b+c}. \tag{3}$$

Let $x := b + c \ge 2$. (3) is written as $$x\left(\frac{72}{\sqrt{25+36x^2} + 5} - \frac{12}{1+x}\right) + \sqrt{25x^2+288x} \ge 5x + \frac{24x}{1+x},$$ or $$\frac{2\sqrt{25 + 36x^2}}{x} + \sqrt{25x^2 + 288x} \ge \frac{10}{x} + 5x + \frac{36x}{1+x}. \tag{4}$$ (4) is true. Note: After twice squaring both sides, (4) is equivalent to $\frac{576(16x^5+193x^4+78x^3+231x^2+56x+50))(x-2)^2}{x(1+x)^4} \ge 0$.

River Li
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