<s>[INST] Maximize $f(x,y)=x+y$ subject to $\sqrt{4-2 x}+\sqrt{4-2 y}=\sqrt{x y}$ Given that $x,y \in [0,2]$ find the maximum value of $x+y$ if $$\sqrt{4-2 x}+\sqrt{4-2 y}=\sqrt{x y}$$
Looking for an elementary approach. No Lagrange multipliers please.
My try:
I tried using CS inequality:
$$\sqrt{4-2x}+\sqrt{4-2y}\leq \sqrt{2}\times \sqrt{8-2(x+y)}$$
$\implies$
$$\sqrt{xy} \le \sqrt{2} \times \sqrt{8-2(x+y)}$$
But not able to proceed?
 [/INST] Remark: Alternatively, we can use the substitutions $\sqrt{4 - 2x} = u, \sqrt{4-2y} = v$.

From $\sqrt{4-2x} + \sqrt{4-2y} = \sqrt{xy}$, we have
$$(8 - 2x - 2y) + 2\sqrt{(4-2x)(4-2y)} \ge xy$$
or
$$4(8 - 2x - 2y) + 8\sqrt{(4-2x)(4-2y)} \ge 4xy$$
or
$$4(8 - 2x - 2y) + 8\sqrt{(4-2x)(4-2y)} \ge (4-2x)(4-2y) - 16 + 8x + 8y$$
or
$$ 64 - 16x - 16y \ge \Big(4 - \sqrt{(4-2x)(4-2y)}\Big)^2$$
or
$$\sqrt{64 - 16x - 16y} \ge 4 - \sqrt{(4-2x)(4-2y)}$$
or
$$\sqrt{64 - 16x - 16y} + \sqrt{(4-2x)(4-2y)} \ge 4.$$
Using $\sqrt{(4-2x)(4-2y)} \le \frac{4-2x + 4 - 2y}{2}$, we have
$$\sqrt{64 - 16x - 16y} + \frac{4-2x + 4 - 2y}{2} \ge 4$$
or
$$\sqrt{64 - 16(x + y)} \ge x + y$$
or
$$64 - 16(x + y) \ge (x + y)^2$$
which results in
$$x + y \le 8\sqrt 2 - 8.$$
Also, if $x = y = 4\sqrt 2 - 4$, we have $x, y\in [0, 2]$
and $\sqrt{4-2x} + \sqrt{4-2y} = \sqrt{xy}$
and $x + y = 8\sqrt 2 - 8$.
Thus, the maximum of $x + y$
is $8\sqrt 2 - 8$.
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