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I'm having trouble showing the following.

If $a>b\ge 0$, then $$\frac{a+b}{2}-\sqrt{ab}<1 \iff \begin{cases}a+b<2, \text{ or,} \\ a+b \ge 2 \text{ and } (a-b)^2 < 4(a+b)-4.\end{cases}.$$

If $a+b<2$, the left-hand side holds trivially. I'm having trouble with the bottom line on the right-hand side.

Here are some attempts. \begin{align*} (a-b)^2 &< 4(a+b)-4\\ (a+b)^2 &< 4(a+b)-4+4ab\\ (a+b)(a+b-4) &< 4ab-4\\ &? \end{align*}

And in the other direction, \begin{align*} a+b &< 2(1+\sqrt{ab})\\ (a+b)^2 &< 4(1+2\sqrt{ab}+ab)\\ (a-b)^2 &< 4(1+2\sqrt{ab}) \le 4+4(a+b) & \text{GM-AM}\\ \end{align*} [so close!]

Any hints would be appreciated. Thanks!

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1 Answer 1

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Hint: $$(a-b)^2 < 4(a+b)-4$$ $$(a+b)^2 < 4(a+b)-4+4ab$$ $$(a+b-2)^2 < 4ab$$ Since $a>b>0$ then $$a+b<2+2\sqrt{ab}$$

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