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I have been attempting to prove the following for $a \geq b \geq c > 0$

$\frac{(a^2-b^2)}{c}+\frac{(c^2-b^2)}{a}+\frac{(a^2-c^2)}{b}\geq 3a-4b+c$

I first identified which terms were definitely positive based on the restrictions. After factoring the numerators and comparing the terms on the left side, I determined that $\frac{(a^2-b^2)}{c}+\frac{(a^2-c^2)}{b}\geq \frac{-(c^2-b^2)}{a}$ since $\frac{(c^2-b^2)}{a}$ is a negative term.

Does anyone have suggestions as to how to proceed with this inequality? I have been trying to compare individual terms, but to no avail. Thanks.

ra1nmaster
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1 Answers1

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Write the main inequality as $f(a,b,c)\geq 0$, then show that $f$ is increasing in $a$ under the assumption of the problem. Therefore, $f(a,b,c)\geq f(b,b,c)=b-c\geq 0$. The equation holds only when $a=b=c$.

S.B.
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  • Can you elaborate on how this could complete the proof? I understand how b - c > 0 was obtained. I am just unsure on how this completes the proof. – ra1nmaster Aug 02 '13 at 22:47
  • Just to be clear I didn't complete your approach, I proved the main inequality itself. We wanted to show $f(a,b,c)\geq 0$ and that's what I showed above. What's specifically not clear? – S.B. Aug 02 '13 at 23:24