Question by Dao Trong Luan - Discuss with MathYouLike

Dao Trong Luan Coordinator

13/12/2017 at 19:54

Given: \(\left\{{}\begin{matrix}x,y>0\\x+y=1\end{matrix}\right.\)

Find the Minimum of \(A=\dfrac{1}{x^2}+\dfrac{1}{y^2}\)

List of answers

Kaya Renger Coordinator 13/12/2017 at 23:12

Applying inequality AM-GM for two non-negative numbers , we have

\(x+y\ge2\sqrt{xy}\)

=> \(xy\le\dfrac{1}{4}\)

Analyze expression A

=> \(A=\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{x^2+y^2}{x^2y^2}\ge\dfrac{2xy}{x^2y^2}=\dfrac{2}{xy}\ge\dfrac{2}{\dfrac{1}{4}}=8\)

So Min_A = 8

Equation occurs when and only when \(x=y=\dfrac{1}{2}\)
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Faded 28/01/2018 at 20:45

Applying inequality AM-GM for two non-negative numbers , we have

$x + y \geq 2 \sqrt{x y}$

=> $x y \leq \frac{1}{4}$

Analyze expression A

=> $A = \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{x^{2} + y^{2}}{x^{2} y^{2}} \geq \frac{2 x y}{x^{2} y^{2}} = \frac{2}{x y} \geq \frac{2}{\frac{1}{4}} = 8$

So Min_A = 8

Equation occurs when and only when $x = y = \frac{1}{2}$

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