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For positive real numbers $a, b, c$, show that

$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}.$$

I don't know how to solve this at all. Can you provide any hints?

Yuxiao Xie
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    Perhaps this might help: http://math.stackexchange.com/questions/883436/prove-fraca2b-fracb2c-fracc2a-geq-3a2b2c2/883490#883490 – Martin Sleziak Apr 17 '16 at 07:10
  • @MartinSleziak Still have to expand the polynomial. Are there any methods that doesn't require expansions like that, since this inequality is homogeneous? – Yuxiao Xie Apr 17 '16 at 07:18

2 Answers2

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We have to prove that $\implies$

$\sum_{cyclic}\frac{a^2}{b}\geq{\frac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ca}}$

$\implies (ab+bc+ca)(\sum_{cyclic}\frac{a^2}{b})\geq(a+b+c)(a^2+b^2+c^2)$

Now expand both sides and many terms are cancelled out.Then we get the

inequality $\implies(\sum_{cyclic}\frac{ca^3}{b})\geq(ac^2+ba^2+cb^2)$

Now use the AM-GM inequality to

1.$(\frac{ca^3}{b}\;,\frac{ab^3}{c})$

2.$(\frac{ab^3}{c}\;,\frac{bc^3}{a})$

3.$(\frac{bc^3}{a}\;,\frac{ca^3}{b})$

Now the required inequality is proved.

sayan
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$$LHS-RHS=\frac{1}{2(ab+bc+ca)}\cdot \left[\frac{b(ab-c^2)^2}{ca}+\frac{c(bc-a^2)^2}{ab}+\frac{a(ca-b^2)^2}{bc}\right]\ge 0$$

BestChoice123
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