<s>[INST] Minimum value of reciprocal squares I am bit stuck at a question. The question is : given: $x + y = 1$,  $x$ and $y$ both are positive numbers. What will be the minimum value of: $$\left(x + \frac{1}{x}\right)^2 + \left(y+\frac{1}{y}\right) ^2$$ 
I know placing $x = y$  will give the right solution. Is there any other solution?
 [/INST] We have $(x+1/x)^2+(y+1/y)^2=x^2+y^2+1/x^2+1/y^2+4=x^2+y^2+\frac{x^2+y^2}{x^2y^2 }+4\geq 1/2+16/2+4=25/2$. Here note that $x+y=1$ implies $xy\leq 1/4$ then $1/(xy)^2\geq 16$ and $x+y=1$ implies $x^2+y^2=1-2xy\geq 1-(x^2+y^2)$ so $x^2+y^2\geq 1/2$. For $x=y=1/2$ we have equality.
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