2

Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality: $$ \frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c} $$

I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.

I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.

I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.

Math Olympian
  • 53

4 Answers4

4

$$\sum{a\over bc}=3$$

$$\sum{a\over b^2c^2}\times(a+b+c)\geq(\sum{a\over bc})^2$$ by Cauchy.

cr001
  • 12,598
  • Can you please elaborate further? – Math Olympian Jan 01 '21 at 11:10
  • @Math Olympian First line comes from dividing both sides by $abc$ from the given. Second line comes from direct application of Cauchy Inequality. – cr001 Jan 01 '21 at 11:12
  • Okay, I follow that! But from where did the second line come? – Math Olympian Jan 01 '21 at 11:14
  • It's the application of Cauchy Inequality. The ${a\over b^2c^2}$ term in the first sum and the $a$ term from the second sum gives the $a\over bc$ term in the right side. You multiply $a$ and $a\over b^2c^2$ and square root the result to obtain $a\over bc$. Then do the similarly for the other two corresponding terms. – cr001 Jan 01 '21 at 11:19
  • Okay! Now I understand, but how does this prove it? – Math Olympian Jan 01 '21 at 11:25
  • @MathOlympian are you taking any effort to understand the proof,it is pretty direct and i feel the solution is already spoonfed.Please make some effort to understand .... – Albus Dumbledore Jan 01 '21 at 11:28
  • Oh wait! Nvm. I got it! Thanks a lot! @cr001 your explanation really helped! – Math Olympian Jan 01 '21 at 11:30
0

Proposed inequality can be rewriten like this:$$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$$

By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^2$$ and we are done!

nonuser
  • 90,026
0

By Cauchy-Schwarz, \begin{align} \left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right)(a+b+c) & \geq \left(\dfrac{a}{bc} + \dfrac{b}{ca} + \dfrac{c}{ab}\right) ^2 \\ & = \left (\dfrac{a^2+b^2+c^2}{abc} \right)^2 \\ & = \left (\dfrac{3abc}{abc} \right)^2 \\ & = 9. \\ \end{align} Thus, dividing both sides of the preceding inequality by $a+b+c$, we have that $\left(\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \right) \geq \dfrac{9}{a+b+c}$, and we are done.

Hello
  • 2,133
-1

[enter image description here][1]

[1]: https://i.stack.imgur.com/VF8FZ.jpg hi this is my answer is it true?

ali..
  • 1