How do I prove $\sqrt{y+c+b} + \sqrt{x+c+b} \geq \sqrt{y+x+c} + \sqrt{c+2b}$ [closed]

I keep trying different combinations for $y,x,b$ and $c$, and I found that the inequality $\sqrt{y+c+b} + \sqrt{x+c+b} \geq \sqrt{y+x+c} + \sqrt{c+2b}$, $\forall$ $y \geq x \geq b>0, c \geq b$ is always true (millions of combinations actually).

I have been working on this these days and I did not found a formal proof. I would be grateful if someone can provide me any hint or help.

Solution 1:

By squaring both sides, we find that the inequality is equivalent to $$(x + y + 2b + 2c) + 2\sqrt{(y+c+b)(x+c+b)} \geq (x + y + 2b + 2c) + 2\sqrt{(y+x+c)(c+2b)}.$$

So, we just need to show that $$(y+c+b)(x+c+b) \geq (y+x+c)(c+2b).$$ To prove that, simply expand both sides, then cancel out the terms to get $$xy + b^2 \geq bx + by.$$ This is equivalent to $$(x-b)(y-b) \geq 0,$$ which is clearly true because we are given that $x \geq b$ and $y \geq b$.