Question

If $x,y,z\in R^{+}$ then $\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{xz}{x+z}$ is always

• A. $\le 2(x+y+z)$
• B. $\le \frac{1}{2}(x+y+z)$
• C. $\le 4(x+y+z)$
• D. None of the above

Answer 1

The correct option is B

$\le \frac{1}{2}(x+y+z)$

From A.M. $\ge$ G.M.$\ge$ H.M.

We know that A.M. $\ge$ H.M. and G.M. $\ge$ H.M.

$\frac{x+y}{2}\ge \frac{2xy}{x+y}$
$\Rightarrow$ $\frac{xy}{x+y}\le \frac{1}{4}(x+y)\cdots \left(1\right)$
$\frac{y+z}{2}\ge \frac{2yz}{y+z}$
$\Rightarrow$ $\frac{yz}{y+z}\le \frac{1}{4}(y+z)\cdots \left(2\right)$

Similarly, $\frac{xz}{x+z}\le \frac{1}{4}(x+z)\cdots \left(3\right)$

Adding (1), (2) and (3),

$\frac{xy}{x+y}+\frac{xy}{x+y}+\frac{xz}{x+z}\le \frac{1}{4}(2x+2y+2z)$

$\Rightarrow \frac{xy}{x+y}+\frac{xy}{x+y}+\frac{xz}{x+z}\le \frac{1}{2}(x+y+z)$

Hence, Option b. is true