Prove that $\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$

Question

$$\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$$ I tried applying a.m. g.m inequality to l.h.s and tried to find upper bound for l.h.s and lower bound for r.h.s but i am not getting answer .

Answer 1

I suppose that $a,b,c,d$ are $>0$. An idea is to put $$F(x)=\frac{(x+c)(b+d)}{x+b+c+d}-\frac{xb}{x+b}-\frac{cd}{c+d}$$ and to compute the derivative: $$F^{\prime}(x)=\frac{(b+d)^2}{(x+b+c+d)^2}-\frac{b^2}{(x+b)^2}$$

This show that the minimum of $F$ on $]0,+\infty[$ is obtained for $x=bc/d$, and to finish you have to compute this minimum.

Answer 2

Not a beautiful proof I have to admit, but it is a proof:

Multiply 2 sides of equation with $(a+b)(c+d)(a+b+c+d)$, we get to the equivalent form: $$ ab(c+d)(a+b+c+d) + cd(a+b)(a+b+c+d) \leq (a+c)(b+d)(a+b)(c+d). $$ Work out both sides of the equation above to reduce it to the following equivalent form: $$ 2abcd \leq a^2d^2 + b^2c^2, $$ which is obviously true due to the AM-GM inequality.

Answer 3

By C-S we obtain: $$\frac{ab}{a+b}+\frac{cd}{c+d}=a+c+\left(\frac{ab}{a+b}-a\right)+\left(\frac{cd}{c+d}-c\right)=$$ $$=a+c-\left(\frac{a^2}{a+b}+\frac{c^2}{c+d}\right)\leq a+c-\frac{(a+c)^2}{a+b+c+d}=\frac{(a+c)(b+d)}{a+b+c+d}.$$ Done!