parmenides51 wrote: The real numbers $x,y,z$ are such that $x+y+z=4$ and $0 \le x,y,z \le 2$. Find the minimun value of the expression $$A=\sqrt{2+x}+\sqrt{2+y}+\sqrt{2+z}+\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$$. The answer is $6+3\sqrt2$ because $$\sqrt{2+x}+\sqrt{y+z}\geq2+\sqrt2.$$Indeed, let $f(x)=\sqrt{2+x}+\sqrt{4-x}.$ Thus, $f$ is a concave function, which says that it's enough to check the last inequality for $x\in\{0,2\}.$ The equality occurs for $(2,2,0)$.

arqady wrote: parmenides51 wrote: The real numbers $x,y,z$ are such that $x+y+z=4$ and $0 \le x,y,z \le 2$. Find the minimun value of the expression $$A=\sqrt{2+x}+\sqrt{2+y}+\sqrt{2+z}+\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$$. The answer is $6+3\sqrt2$ because $$\sqrt{2+x}+\sqrt{y+z}\geq2+\sqrt2.$$The equality occurs for $(2,2,0)$. How did you get $\sqrt{2+x}+\sqrt{y+z}\geq 2+2\sqrt{2}$ ? Did you use differentiation or some method as mine ?

parmenides51 wrote: The real numbers $x,y,z$ are such that $x+y+z=4$ and $0 \le x,y,z \le 2$. Find the minimun value of the expression $$A=\sqrt{2+x}+\sqrt{2+y}+\sqrt{2+z}+\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$$. Notice: $$ \begin{aligned} \left( \sqrt{x + 2} + \sqrt{4 - x} \right)^2 &= 2 + x + 4 - x + 2\sqrt{8 + 2x - x^2} \\ &= 6 + 2 \sqrt{8 + x(2 - x)} \\ &\geq 6 + 4\sqrt2 = \left( 2 + \sqrt2 \right)^2 \end{aligned} $$Therefore, $\boxed{A \geq 6 + 3 \sqrt2}$. Equality occurs if two of $x$, $y$, $z$ are $2$ and the other one is $0$.

parmenides51 wrote: The real numbers $x,y,z$ are such that $x+y+z=4$ and $0 \le x,y,z \le 2$. Find the minimun value of the expression $$A=\sqrt{2+x}+\sqrt{2+y}+\sqrt{2+z}+\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$$.

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