Question

Use HM $\le$ AM and find maximum value of $\frac{xy}{x+y}$ + $\frac{yz}{y+z}$ + $\frac{xz}{x+z}$

• A.
• B. x + y + z
• C.
• D. 2(xy + yz + xz)

Answer 1

The correct option is C

From $AM\ge GM\ge HM$

We know that AM $\ge$ HM and GM $\ge$ HM

(i) $\Rightarrow$ $\frac{x+y}{2}$ $\ge$ $\frac{2xy}{x+y}$ $\Rightarrow$ $\frac{xy}{x+y}$ $\le$ $\frac{1}{4}$(x+y) --------(1)

$\frac{y+z}{2}$ $\ge$ $\frac{2yz}{y+z}$ $\Rightarrow$ $\frac{yz}{y+z}$ $\le$ $\frac{1}{4}$(y + z) --------(2)

Similarly $\frac{xz}{x+z}$ $\le$ $\frac{1}{4}$(x + z) --------------------(3)

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

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

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