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Let $a,b,c>0$ with $a^2+b^2+c^2=3$.

Prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{2}{abc}\geq1.$$

I can prove it by using "mixing variables": \begin{align*} P&=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{2}{a b c} \\ &\geq \frac{2}{a b}+\frac{1}{c^2}-\frac{2}{a b c} \\ &=\frac{2}{a b}\left(1-\frac{1}{c}\right)+\frac{1}{c^2} \\ &\geq \frac{4}{a^2+b^2}\left(1-\frac{1}{c}\right)+\frac{1}{c^2} \\ &=\frac{4}{3-c^2}\left(1-\frac{1}{c}\right)+\frac{1}{c^2} \\ &=f(c). \end{align*}

WLOG $c=\max\{a,b,c\}$, and it is very easy to prove. I don't know if there is any other solution? If yes please show me (I'm thinking of using mixing variables with $f(abc)$).

kipf
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trungbk
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  • Another way is to maybe apply AM-GM to get

    $$\frac1{a^2} + \frac1{b^2} + \frac1{c^2} \leq \frac{3}{abc},$$

    and then show that $abc \leq 1$ using AM-GM again.

     – eeen Jun 24 '23 at 16:42
  • How do you prove the above inequality? – trungbk Jun 24 '23 at 17:14

3 Answers3

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pqr method:

Let $x = a^2, y = b^2, z = c^2$. Then $x + y + z = 3$. We need to prove that $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - \frac{2}{\sqrt{xyz}} \ge 1. \tag{1}$$

Let $p = x + y + z = 3, q = xy + yz + zx, r = xyz $.

(1) is written as $$\frac{q}{r} - \frac{2}{\sqrt r} \ge 1.$$

Using $q^2 \ge 3pr$, it suffices to prove that $$\frac{\sqrt{3pr}}{r} - \frac{2}{\sqrt r} \ge 1$$ or $$1 \ge \sqrt r$$ which is true using $r \le (p/3)^3 = 1$.

We are done.

River Li
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Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(w^3)\geq0,$ where $f$ is a concave function, which says that it's enough to prove our inequality for an extremal value of $w^3,$ which by $uvw$ says, it remains to check one case only( for $w^3\rightarrow0^+$ it's obvious): an equality case of two variables and the rest is smooth:

Let $b=a$ and $c=\sqrt{3-2a^2},$ where $0<a<\sqrt{\frac{3}{2}}.$ thus, we need to prove that: $$\frac{2}{a^2}+\frac{1}{3-2a^2}\geq1+\frac{2}{a^2\sqrt{3-2a^2}}$$ or $$a^4-3a^2+3\geq\sqrt{3-2a^2},$$ which is true by AM-GM: $$a^4-3a^2+3=\frac{(3-2a^2)^2+3}{4}\geq\frac{4\sqrt[4]{(3-2a^2)^2\cdot1^3}}{4}=\sqrt{3-2a^2}$$ and we are done.

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Michael Rozenberg
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Another way.

We need to prove $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{2}{\sqrt{abc}}\geq1,$$ where $a$, $b$ and $c$ are positives such that $a+b+c=3$ or $$ab+ac+bc-abc\geq2\sqrt{abc}$$ or $$(a+b+c)(ab+ac+bc)-3abc\geq2\sqrt{\frac{1}{3}abc(a+b+c)^3}$$ or $$3\left(\sum_{cyc}(a^2b+a^2c)\right)^2\geq4abc(a+b+c)^3,$$ which is by AM-GM and P-M(or Jensen for $f(x)=x^{1.5}$): $$3\left(\sum_{cyc}(a^2b+a^2c)\right)^2\geq3\left(\sum_{cyc}(2a^2\sqrt{bc})\right)^2=12abc\left(\sum_{cyc}\sqrt{a^3}\right)^2\geq4abc(a+b+c)^3.$$

Michael Rozenberg
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