<s>[INST] $a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge\frac{100}{3}$ $a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that

$$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\geq\frac{100}{3}$$

 [/INST] The function $f(x) =(x+x^{-1} )^2 $ is convex on $(0,\infty) $ hence $$ \frac{1}{3} (f(a) +f(b) +f(c))\geq f(\frac{a+b+c}{3}) =(3+3^{-1} )^2 $$
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