Prove that $\sqrt{x^2+y^2}+\left(2-\sqrt{2}\right) \sqrt{xy} \geq x+y$ if $x$ and $y$ are real positive numbers!

Question

Prove that $$\sqrt{x^2+y^2}+\left(2-\sqrt{2}\right) \sqrt{xy} \geq x+y$$ if $x$ and $y$ are real positive numbers!

Since both sides of the inequality are positive, when we square it, we get $$x^2+y^2+2 \sqrt{x^2+y^2} (2-\sqrt{2}) \sqrt{xy}+(4-4 \sqrt{2}+2)xy \geq x^2+2xy+y^2,$$ $$ 2 \sqrt{x^2+y^2} (2-\sqrt{2 }) \sqrt{xy} \geq 4 (\sqrt{2}-1)xy.$$

Dividing both sides of the inequality by $$2 \sqrt{xy}>0$$ and then squaring both sides of the inequality (both sides are positive), we gradually obtain $$\sqrt{x^2+y^2} \cdot \left(2-\sqrt{2}\right) \geq 2 \left(\sqrt{2}-1\right) \sqrt{xy},$$ $$ \left(x^2+y^2\right) \cdot \left(6-4 \sqrt{2}\right) \geq 4 \left(3-2 \sqrt{2}\right)xy.$$

Dividing both sides of the inequality by $$\left(6-4\sqrt{2}\right) >0,$$ we get $$x^2+y^2 \geq 2xy$$ or $$\left(x-y\right) ^2 \geq 0.$$ Since the square of the number is non-negative, the last inequality is true. Since equivalent transformations were performed, the given inequality is also true for all real numbers $x$.

Is there any shorter way to solve this problem? I will be very very grateful for any hint if I need to use any auxiliary inequality as Cauchy or something else! Thanks!

Answer 1

Based on Post #$5$ of the AoPS thread Inspired by Poland MO, we have

$$\begin{equation}\begin{aligned} x^2 + y^2 & \ge 2xy \\ (x^2 + y^2)xy & \ge 2x^2y^2 \\ \sqrt{(x^2 + y^2)xy} & \ge \sqrt{2}xy \end{aligned}\end{equation}$$

Thus, from the LHS after your initially squaring, we then get

$$\begin{equation}\begin{aligned} & x^2+y^2+2\sqrt{x^2+y^2}(2-\sqrt{2})\sqrt{xy}+(4-4\sqrt{2}+2)xy \\ & = x^2+y^2+2(\sqrt{(x^2+y^2)xy})(2-\sqrt{2})+(6-4\sqrt{2})xy \\ & \ge x^2 + y^2 + 2(\sqrt{2}xy)(2-\sqrt{2}) + (6-4\sqrt{2})xy \\ & = x^2 + y^2 + (4\sqrt{2} - 4)xy + (6-4\sqrt{2})xy \\ & = x^2 + y^2 + 2xy \\ & = (x + y)^2 \end{aligned}\end{equation}$$

Answer 2

Let $\sqrt{\dfrac{x^2+y^2}{2}} = A, C = \sqrt{xy}$ and $B = \dfrac{x+y}{2}.$ Then you have $A\geq B\geq C$ and $A^2+C^2 = 2B^2$ and so $\sqrt{2}B\geq A.$ We want to prove: $$A + (\sqrt{2}-1)C\geq \sqrt{2}B\iff A-C\geq \sqrt{2}(B-C).$$ But $$\dfrac{A-C}{B-C} = \dfrac{A^2-C^2}{(B-C)(A+C)} = \dfrac{2(B^2-C^2)}{(B-C)(A+C)} = \dfrac{2(B+C)}{A+C}\geq\dfrac{\sqrt{2}A + \sqrt{2}C}{A+C} = \sqrt{2}.$$

Of course the above assumes $x\neq y$, which ensures $A>B>C.$ The problem is trivial when $x = y.$