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

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

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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.

Answer 1

Write $b^2+c^2=a^2r^2$, $r\in [0,\,1]$ so $A=r^2+\frac{a^4 r^2}{(bc)^2}$. Write $b=arC,\,c=arS$ with $C=\cos t,\,S=\sin t$ (say) so $A=r^2+\frac{1}{r^2 C^2 S^2}\ge r^2+\frac{4}{r^2}$, with equality iff $\sin 2t=\pm 1,\,C^2=S^2=\frac{1}{2}$. Since $\frac{d}{dr}(r^2+\frac{4}{r^2})=2r(1-\frac{4}{r^4})<0$ for $r\in [0,\,1]$, we obtain the minimum at $r=1$. Thus $A$ has minimum $5$, obtained with $b^2=c^2=\frac{a^2}{2}$ as you thought.

Answer 2

Hint: we get $$\frac{1}{a^2}(b^2+c^2)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\geq 5$$ The equal sign holds, if $$a=1,b=c=\frac{1}{\sqrt{2}}$$

Answer 3

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!