TITLE: Find the minimize of $A=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)$
QUESTION [0 upvotes]: Let $a,b,c>0$ such that $b^2+c^2\le a^2$. Find the minimize of $$A=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)$$

My try: By AM-GM: $LHS=\frac{a^2}{b^2}+\frac{b^2}{a}+\frac{a^2}{c^2}+\frac{c^2}{a^2}$
$=\frac{a^2}{b^2}+4\frac{b^2}{a}+\frac{a^2}{c^2}+4\frac{c^2}{a^2}-\frac{3\left(b^2+c^2\right)}{a^2}$
$\ge 2\sqrt{\frac{a^2}{b^2}\cdot 4\frac{b^2}{a}}+2\sqrt{\frac{a^2}{c^2}\cdot 4\frac{c^2}{a^2}}-\frac{3\cdot a^2}{a^2}$  
$\ge 4+4-3=5$
$"="\Leftrightarrow b=c=\frac{a}{\sqrt{2}}$
My idea started by the equal occurs when $b=c=\frac{a}{\sqrt{2}}$ but if I don't know the equal occurs when $b=c=\frac{a}{\sqrt{2}}$ i can't sol it. Help me sol it.

REPLY [1 votes]: The first thing, which we need to check, when we want to prove inequality it's to check the equality case, which you made. It says that your solution is beautiful!
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