# Solve an inequality using Cauchy-Schwarz Inequality

Le $a,b,c,d \in \mathbb{R^{+}}$. Using Cauchy-Schwarz Inequality prove that the following inequality holds:

$$\frac{1}{\frac{1}{a+c} + \frac{1}{b+d}} \ge \frac{1}{\frac 1a + \frac 1b} + \frac{1}{\frac 1c + \frac 1d}$$

I've been trying for a while, but I haven't been able to make any progress. Actually I have solved this ineqaulity using 2 methods, but both make no use of the Cauchy-Schwarz inequality. In the first solution I clear the denominators and it reduces to a rather simple inequality. The original inequality is equivalent to:

$$\frac{(a+c)(b+d)}{a+b+c+d} \ge \frac{ab}{a+b} + \frac{cd}{c+d} = \frac{ab(c+d) + cd(a+b)}{(a+b)(c+d)}$$

After clearing the denominators, multiplying and canceling everything out we are left with an inequality that's true from AM-GM:

$$b^2c^2 + a^2d^2 \ge 2abcd$$

The other solution makes use of the Minkowski Inequality. Substitute $r=1; s=-1; a_{11}=a; a_{12}=b; a_{21}=c; a_{22}=d; m=n=2$ in the following form of the Minkowski Inequality and we get the original inequality.

$$\left(\sum^{m}_{j=1}\left(\sum^{n}_{i=1} a_{ij}^r\right)^{\frac sr}\right)^{\frac 1s} \ge \left(\sum^{n}_{i=1}\left(\sum^{m}_{j=1} a_{ij}^s\right)^{\frac rs}\right)^{\frac 1r}$$

Starting from the equivalent inequality $$\frac{(a+c)(b+d)}{a+b+c+d} \ge \frac{ab}{a+b} + \frac{cd}{c+d} = \frac{ab(c+d) + cd(a+b)}{(a+b)(c+d)}$$ and cross-multiplying results in \begin{align}(a+b)(a+c)(b+d)(c+d)&\geq\{(a+b)+(c+d)\}\{ab(c+d)+cd(a+b)\}\\&=cd(a+b)^2+ab(c+d)^2+(a+b)(c+d)(ab+cd)\end{align} which (after subtracting $(a+b)(c+d)(ab+cd)$ from both sides) simplifies to $$(a+b)(c+d)(bc+ad)\geq cd(a+b)^2+ab(c+d)^2$$ completing the square on the right hand side, and rearranging the cross term of the square, leads to the inequality being expressed as follows $$(a+b)(c+d)(bc+ad+2\sqrt{abcd})\geq \{\sqrt{cd}(a+b)+\sqrt{ab}(c+d)\}^2\\\Rightarrow(a+b)(c+d)(\sqrt{ad}+\sqrt{bc})^2\geq\{\sqrt{cd}(a+b)+\sqrt{ab}(c+d)\}^2$$ Dividing both sides by positive value $(\sqrt{ad}+\sqrt{bc})^2$ leaves the direction of the inequality unchanged, resulting in $$(a+b)(c+d)\geq\left[\frac{\sqrt{cd}(a+b)+\sqrt{ab}(c+d)}{\sqrt{ad}+\sqrt{bc}}\right]^2\\=(\sqrt{ac}+\sqrt{bd})^2$$ Thus, we end up with the following inequality $$(a+b)(c+d)\geq(\sqrt{ac}+\sqrt{bd})^2$$ which can be seen to be the Cauchy-Schwarz inequality, $$\sum_{j=1}^N|x_j|^2\sum_{k=1}^N|y_k|^2\geq\left|\sum_{i=1}^Nx_i\bar{y_i}\right|^2$$ with $x_1=\sqrt{a},x_2=\sqrt{b}$ and $y_1=\sqrt{c},y_2=\sqrt{d}$