Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$

Question

For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$

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I think if we let $x=\tan A;y=\tan B;z=\tan C$, then

$$P\Leftrightarrow \displaystyle \text {sin}A+\text {sin}B+ \text{sin}C\leq \frac{3}{2}. \hspace{2cm}(1)$$

But I can't prove $(1)$.

Answer 1

For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3\sqrt3}{2}$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$\sum_{cyc}\frac{1}{\sqrt{x^2+xy+xz+yz}}\leq\frac{3\sqrt3}{2\sqrt{xy+xz+yz}}$$ or $$\sum_{cyc}\sqrt{x+y}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}.$$ But by C-S $$\left(\sum_{cyc}\sqrt{x+y}\right)^2\leq(1+1+1)\sum_{cyc}(x+y)=6(x+y+z).$$

Thus, it remains to prove that $$\sqrt{6(x+y+z)}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}$$ or $$9(x+y)(x+z)(y+z)\geq8(x+y+z)(xy+xz+yz)$$ or $$\sum_{cyc}z(x-y)^2\geq0.$$ Done!

Answer 2

Hint : You're making a mistake.

You have to prove this $$ \cos A + \cos B + \cos C \le \frac{3}{2} $$ because : $$ 1+\tan^2 \theta = \sec^2 \theta $$

Answer 3

Like How to show that the triangle is equilateral triangle?,

$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$ with $A+B+C=\pi$

Now use this