Problem in solving an inequality.

Question

Let $a,b,c,d \in \Bbb R^{+}$ be such that $a>b$ and $c>d$ with $a+b=c+d.$ If $a > c$ (and consequently $b<d$) then show that

$$\sqrt a + \sqrt b < \sqrt c + \sqrt d.$$

My attempt $:$

Suppose on the contrary we have

$$\sqrt a + \sqrt b \geq \sqrt c + \sqrt d.$$ Squaring both sides we have $$a+b + 2\sqrt {ab} = c+d + 2 \sqrt {cd}.$$ Since $a+b= c+d$ we have $$\sqrt {ab} \geq \sqrt {cd}.$$ Again squaring we have $$ab \geq cd.$$ Now how do I draw a contradiction from here? I note that If $b \geq c$ then $a+b > 2c$ since it is given that $a>c.$ Now since $c > d$ so we have $a+b > c+d,$ a contradiction to the fact that $a+b=c+d.$ So we must have $b < c.$ $b<d$ as well for the same reason. Hence we can conclude that $a>c>d>b.$ Does it help me anyway in proving the required inequality?

Any help will be highly appreciated. Thank you very much.

Answer 1

Hint: $(a+b)^2-(a-b)^2=4ab$

Can you see how you can use this standard identity together with what you already know?

Answer 2

$a-c=d-b$ so $(\sqrt a-\sqrt c)(\sqrt a+\sqrt c)=(\sqrt d-\sqrt b)(\sqrt d+\sqrt b)$.

Since $a>b$, $c>d$, we have $$\sqrt a+\sqrt c>\sqrt b+\sqrt d.$$ Hence $$\sqrt a-\sqrt c<\sqrt d-\sqrt b,$$ as desired.