# Inequality in Convex Quadrilateral

### Problem

### Solution

Let $a+b+c+d=2s.$ Then $b+c+d-a=2(s-a),$ $a+c+d-b=2(s-b),$ $a+b+d-c=2(s-c),$ $a+b+c-d=2(s-d).$ Set $s-a=x,$ $s-b=y,$ $s-c=z,$ $s-d=t.$ Obviously,

$\displaystyle \sum_{cycl}x=\sum_{cycl}(s-a)=\sum_{cycl}a.$

Also

$\displaystyle \sum_{cycl}(s-a)^2=a^2+b^2+c^2+d^2.$

The required inequality can be rewritten as

$(x^2+y^2+z^2+t^2)(\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t})^2\ge (x+y+z+t)^3$

which is the same as

$\displaystyle \left(\sum_{cycl}x^2\right)\left(\sum_{cycl}\sqrt{x}\right)\left(\sum_{cycl}\sqrt{x}\right)\ge (x+y+z+t)^3$

which is a form of *Hölder's inequality*.

### Acknowledgment

The problem, due to Lorian Saceanu, was kindly sent to me by Leo Giugiuc, along with a solution of his. The problem has originally appeared at Saceanu's facebook group Easy and Beautiful Math.

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