Let $a,b, c$ are positive real numbers satisfying the condition $a^2 + b^2 + c^2=6.$ Prove that$$\sqrt{4 -a^2} +\sqrt{4 -b^2} +\sqrt{4 -c^2}\leq 3\sqrt 2$$https://artofproblemsolving.com/community/c6h558229p3246219: Let $a,b, c$ be nonnegative real numbers such that $a^2 + b^2 + c^2=10.$ Prove that$$\sqrt 6+\sqrt 2\leq \sqrt{6-a^2}+\sqrt{6-b^2}+\sqrt{6-c^2}\leq 2\sqrt 6$$

sqing wrote: Let $a,b, c$ are positive real numbers satisfying the condition $a^2 + b^2 + c^2=6.$ Prove that$$\sqrt{4 -a^2} +\sqrt{4 -b^2} +\sqrt{4 -c^2}\leq 3\sqrt 2$$ $$\sqrt{4 -a^2} +\sqrt{4 -b^2} +\sqrt{4 -c^2}\leq\sqrt{3(4 -a^2+4 -b^2 +4 -c^2)}=3\sqrt 2$$

Let's find the minimum value. By setting $x = \sqrt{4 - a^2}, y = \sqrt{4 - b^2}, z = \sqrt{4 - c^2}$ the problem transforms into finding the minimum of $x + y + z$ subject to $x^2 + y^2 + z^2 = 6$ and $0 \leq x, y, z \leq 2$. Without loss of generality, assume $x = \max \{x, y, z\}$, so $x \in \left[ \sqrt 2, 2 \right]$. Since $y, z$ are non-negative, \[ 6 = x^2 + y^2 + z^2 \leq x^2 + (y + z)^2 \]and thus \[ x + y + z \geq x + \sqrt{6 - x^2} \]We now have \[ \left(x + \sqrt{6 - x^2} \right)^2 = 6 + 2\sqrt{x^2(6 - x^2)} = 6 + 2\sqrt{8 + (x^2 - 2)(4 - x^2)} \geq 6 + 4\sqrt 2 = \left(2 + \sqrt 2 \right)^2 \] Therefore $x + y + z \geq 2 + \sqrt 2$. The inequality occurs for example when $x = 2, y = \sqrt 2, z = 0$ so this is indeed the minimum.

Let $a,b, c$ are positive real numbers satisfying the condition $a^2 + b^2 + c^2=6.$ Prove that$$2 + \sqrt 2 \leq\sqrt{4 -a^2} +\sqrt{4 -b^2} +\sqrt{4 -c^2}\leq 3\sqrt 2$$