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If $a,b,c,d$ are positive integers and $a+b+c+d=16$, prove that $$\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2 \geq \frac{289}{4}.$$

I know I have to use some inequality, not sure AM GM will work here or Minkowski inequality. But I only need hints, not a complete solution. I want to work on it myself.

Please provide only Hints.

Michael Rozenberg
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Siddhi V Iyer
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  • Hm, thinking out loud. $289 = (16+1)^2.$ More thinking... –  Mar 14 '12 at 02:46
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    You have two copies of the same expression up there. I'd try and see what I could say about the behavior of $(x+\frac{1}{y})^2+(y+\frac{1}{x})^2$ if $x+y=8$, or more generally $x+y$ is equal to some constant $n$. – KReiser Mar 14 '12 at 02:52

2 Answers2

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First apply the Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality (from http://tinyurl.com/84o57u4)

$$\sqrt{\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4}} \geq \frac{\left(a+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)+\left(b+\frac{1}{d}\right)+\left(d+\frac{1}{b}\right)}{4}$$

Square both sides

$$\frac{\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2}{4} \geq \frac{\left( (a+b+c+d)+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}) \right)^2}{16}$$

Since we already know the value of $a+b+c+d=16$, apply Arithmetic Mean, Harmonic Mean inequality, i.e.

$$ \frac{a+b+c+d}{4} \geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}$$

And take it from there, since you said you only need hints.

Kirthi Raman
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Since $f(x)=x^2+\frac{1}{x^2}$ is a convex function, by Jensen and AM-GM we obtain: $$\sum_{cyc}\left(a+\frac{1}{c}\right)^2=\sum_{cyc}f(a)+2\sum_{cyc}\frac{a}{c}\geq 4f\left(\frac{a+b+c+d}{4}\right)+8=\frac{289}{4}$$

Michael Rozenberg
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