Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$

Question

I found this inequality using unusual calculations in maths Olympics and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.

Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Prove that $$\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$$

Answer 1

By applying the generalized mean inequality twice,

$$\left(\frac{\sqrt{x + y - x^2} + \sqrt{y + z - y^2} + \sqrt{z + w - z^2} + \sqrt{w + x - w^2}}{4}\right)^2\\ \le \frac{1}{4}(2x + 2y + 2z + 2w - x^2 - y^2 - z^2 - w^2)\\ = \frac{1}{2} - \frac{x^2 + y^2 + z^2 + w^2}{4}\\ \le \frac{1}{2} - \left(\frac{x + y + z + w}{4}\right)^2\\ =\frac{1}{2} - \frac{1}{16}\\ = \frac{7}{16}$$

so that

$$\sqrt{x + y - x^2} + \sqrt{y + z - y^2} + \sqrt{z + w - z^2} + \sqrt{w + x - w^2}\le\sqrt7 $$

with equality at $$x = y = z = w = \frac{1}{4}$$

Answer 2

It depends on what you consider to be elementary.

By cauchy-schwarz:

$$ \sum\limits_{cyc} \sqrt{x+y-x^2} \leq2\sqrt{\sum\limits_{cyc}x+y-x^2}=2\sqrt{\frac{7}{4}+\frac{1}{4}(x+y+w+z)^2-(x^2+y^2+w^2+z^2)}.$$

Now, by jenson we have that $$(x+y+w+z)^2\leq 4(x^2+y^2+z^2+w^2). $$

Hence we conclude that $$ \sum\limits_{cyc} \sqrt{x+y-x^2} \leq \sqrt{7}.$$

Edit: I would consider this elementary since jensen and cauchy-schwarz are some of the first inequalities one learns after AM-GM. Moreover, we have equality in the above if and only if$x=y=w=z=1/4.$

Answer 3

By Jensen $\sum\limits_{cyc}\sqrt{x+y-x^2}\geq2\sqrt{\sum\limits_{cyc}(2x-x^2)}=2\sqrt{2-\sum\limits_{cyc}x^2}\leq2\sqrt{2-\frac{1}{4}}=\sqrt7<4\sqrt2-3$