Proving an inequality involving integrals

Let $0<a<1$, $0<b<1$, $c>0$ and $d>0$, prove the following inequality: $$\frac{1}{\frac{1}{a}+\frac{1}{b}}\geq \int_{0}^\infty\frac{1}{\frac{1}{ac\exp(-cx)}+\frac{1}{bd\exp(-dx)}}$$ This one I simply don't know where to start. I figured out that when $c=d$, the inequality becomes: $$\frac{1}{\frac{1}{a}+\frac{1}{b}}= \frac{1}{\frac{1}{a}+\frac{1}{b}}\int_{0}^\infty c\exp(-cx)$$ and the integral becomes one as it is the pdf of an exponential distribution. I also tried Jensen's inequality and HM-GM-AM inequality but it does not seem to work. I'd be really grateful if someone can give me some hints on how to prove this inequality or tell me where to start with.

Many thanks.

Thanks for the interesting problem. We can start with the Cauchy-Schwarz inequality. For positive $k_1$, $k_2$, $y_1$, and $y_2$ $$(k_1y_1 + k_2 y_2)\left(\frac{k_1}{y_1} + \frac{k_2}{y_2}\right) \ge (k_1 + k_2)^2. \qquad (1)$$ This inequality can be rewritten as $$\frac{ k_1 y_1 + k_2 y_2 } { (k_1 + k_2)^2 } \ge \frac{ 1 } { \frac{k_1}{y_1} + \frac{k_2}{y_2} }. \qquad (2)$$ For the inequality stated in the problem, we can use (2) with $k_1 = 1/a$, $k_2 = 1/b$, $y_1 = c \exp(-cx)$, $y_2 = d \exp(-dx)$, and integrate both sides over $x$, and notice that $\int y_1 \, dx = \int y_2 \, dx = 1$, $$\frac{1}{k_1 + k_2} \ge \int \frac{ 1 } { \frac{k_1}{y_1} + \frac{k_2}{y_2} } dx.$$
• Thanks very much for the elegant solution! By the way the r.h.s. of your equation (1) should be $（k_1+k_2)^2 y_1 y_2$. This is way easier than the solution I'm attempting. – user46666 Jan 5 '16 at 9:19