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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?

Ishan Arora
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4 Answers4

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$f(t) = (t + \frac1t)^2 = t^2 + 2 + \frac1{t^2}$ is a sum of convex functions, hence convex; by Jensen's inequality, $f(x)+f(y)\ge 2f(\frac{x+y}2) = 2f(\frac12)$, with equality when $x=y$.

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$x+y=1,~$ x and y both are positive numbers.

Then $x=\sin^2t$ and $y=\cos^2t$ is a valid parameterization, wouldn't you agree ?

What will be the minimum value of $~\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2~$ ?

Hint: Rewrite the expression in question in terms of t, and ask yourself what techniques you already know for finding the extrema $($minima and maxima$)$ of functions of a single variable.

Lucian
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  • This only gives the minimum value of $t$. Why does that necessarily give the minimum value of $x$ and $y$? – beep-boop Aug 02 '15 at 23:57
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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.

zoli
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Euler88 ...
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$(x+1+y/x)^2+ (y+1+x/y)^2 >= 1/2 (x+1+y/x+y+1+x/y)^2 >= 1/2(2+1+x/y+y/x)^2 >=25/2 $

Equality holds when $x/y=y/x$ or $x=y$

Rigel
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