Find maximize $A=\sum_{cyc}\sqrt{1+x^2}+2(\sum_{cyc}\sqrt{x})$

Question

Given $x,y,z$ are positive number satisfy $x+y+z\le 3$. Find the value of maximize $$\sqrt{1+x^{2}}+\sqrt{1+y^{2}}+\sqrt{1+z^{2}}+2(\sqrt{x}+\sqrt{y}+\sqrt{z})$$

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$\sum (\sqrt{1+x^{2}}+\sqrt{2x})\leq \sum \sqrt{2(x^{2}+1+2x)}=\sqrt{2}\sum (x+1)\leq 6 \sqrt{2}$

$(2-\sqrt{2})(\sqrt{x}+\sqrt{y}+\sqrt{z})\leq (2-\sqrt{2})(\sqrt{3(x+y+z)})\leq 3.(2-\sqrt{2})$

$\Rightarrow A\leq 6 \sqrt{2}+3.(2-\sqrt{2})=6+3\sqrt{2}$

i need another way

Answer 1

We need prove $\sqrt{x^2+1}+2\sqrt{x}\le \frac{2+\sqrt{2}}{2}\left(x+1\right)$

$\Leftrightarrow \left(\sqrt{x^2+1}+2\sqrt{x}\right)^2\le \frac{3+2\sqrt{2}}{2}\left(x+1\right)^2$

$\Leftrightarrow \frac{1+2\sqrt{2}}{2}\left(x^2+1\right)-4\sqrt{x\left(x^2+1\right)}+\left(2\sqrt{2}-1\right)x\ge 0$

$\Leftrightarrow \left(\sqrt{x^2+1}-\sqrt{2x}\right)\left(\frac{1+2\sqrt{2}}{2}\sqrt{x^2+1}-\frac{4-\sqrt{2}}{2}\sqrt{x}\right)\ge 0$

Right because $x^2+1\ge 2x$

So $\text{∑}\sqrt{x^2+1}+2\sqrt{x}\le \text{∑}\frac{2+\sqrt{2}}{2}\left(x+1\right)\le 6+3\sqrt{2}$

Answer 2

Let $$f(x) = \sqrt{1+x^2}+2\sqrt{x}$$ For $x>0$, we can show $f(x)$ is convex downward:

$$\frac{df}{dx} = \frac{x}{\sqrt{1+x^2}}+\frac{1}{\sqrt{x}}\\ \frac{d^2f}{dx^2} = \left(1+x^2 \right)^{-3/2} -\frac1{2x^{3/2}} $$ but for $x>0$ we have $1+x^2> 2x > \sqrt[3]{4} x$ so $$ \left(1+x^2 \right)^{-3/2} < (\sqrt[3]{4} x)^{3/2} $$ meaning $f(x)$ is concave downward.

Thus the average of three values of $f(x$) is less than or equal to $f(x)$ at the average value. But by the constraint, the average value of $x,y,z$ is $1$ so $$ \sum_{cyc} f(x) \leq 3f(1) = 6+3\sqrt{2} $$

Answer 3

For $x=y=z=1$ we get the value $6+3\sqrt2$.

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

Indeed, we need to prove that $$\sum_{cyc}(2+\sqrt2-2\sqrt{x}-\sqrt{1+x^2})\geq0$$ or $$\sum_{cyc}\left(\frac{2(1-x)}{1+\sqrt{x}}+\frac{1-x^2}{\sqrt2+\sqrt{1+x^2}}\right)\geq0$$ or $$\sum_{cyc}(x-1)\left(1+\frac{1}{\sqrt2}-\frac{2}{1+\sqrt{x}}-\frac{x+1}{\sqrt2+\sqrt{1+x^2}}\right)+\left(1+\frac{1}{\sqrt2}\right)(3-x-y-z)\geq0$$ or $$\sum_{cyc}(x-1)^2\left(\frac{1}{(1+\sqrt{x})^2}-\frac{x+1}{\sqrt2(\sqrt2+\sqrt{1+x^2})(\sqrt2x+\sqrt{1+x^2})}\right)+$$ $$+\left(1+\frac{1}{\sqrt2}\right)(3-x-y-z)\geq0,$$ for which is's enough to prove that $$\sqrt2(\sqrt2+\sqrt{1+x^2})(\sqrt2x+\sqrt{1+x^2})\geq(x+1)(1+\sqrt{x})^2.$$ Let $1+x=2k\sqrt{x}$.

Hence, by C-S $$\sqrt2(\sqrt2+\sqrt{1+x^2})(\sqrt2x+\sqrt{1+x^2})=$$ $$=\sqrt2\left(\sqrt2+\frac{1}{\sqrt2}\sqrt{2(1+x^2)}\right)\left(\sqrt2x+\frac{1}{\sqrt2}\sqrt{2(1+x^2)}\right)\geq$$ $$\geq\sqrt2\left(\sqrt2+\frac{1+x}{\sqrt2}\right)\left(\sqrt2x+\frac{1+x}{\sqrt2}\right)=$$ $$=\frac{1}{\sqrt2}(3+x)(3x+1)=\frac{1}{\sqrt2}(3x^2+10x+3)=$$ $$=\frac{1}{\sqrt2}(3(4k^2-2)x+10x)=2\sqrt2x(3k^2+1).$$ In another hand, $$(x+1)(1+\sqrt{x})^2=(x+1)(x+1+2\sqrt{x})=2k(2k+2)x=4k(k+1)x.$$ Id est, it remains to prove that $$3k^2+1\geq\sqrt2k(k+1)$$ or $$(3-\sqrt2)k^2-\sqrt2k+1\geq0,$$ which is obvious by AM-GM: $$(3-\sqrt2)k^2-\sqrt2k+1\geq\left(2\sqrt{3-\sqrt2}-\sqrt2\right)k\geq0.$$ Done!