Prove for all positive real numbers $x$ and $y$, if $x+y \le (4xy)/(x+y)$, then $x=y$

Question

Would a direct or indirect proof be appropriate for this theorem?

Answer 1

Hint: Multiplying by $$x+y>0$$ we get $$(x+y)^2\le 4xy$$ can you finish?

Answer 2

Since $x+y>0$, then we obtain

$$ (x+y)^2 \leq 4xy $$

as was pointed out by the Doc

Therefore,

$$ x^2 + 2xy + y^2 \leq 4xy $$

which implies that

$$ x^2+y^2 \leq 2xy $$

which implies that

$$ x^2-2xy +y^2 \leq 0 $$

and so we get

$$ 0 \leq (x-y)^2 \leq 0$$

$x-y$ better be $0$