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If $a, b \in \mathbb{R}_+$ such that $a \neq b$ show that $\sqrt{ab} < \frac{a+b}{2}$

Here is my attempt:

Proof: Suppose that $\sqrt{ab} \geq \frac{a+b}{2}$ , since both sides of the inequality are positive, it follows that: $4ab \geq a^2 +2ab + b^2 \Rightarrow (a-b)^2 \leq 0$, note that the expression $(a-b)^2<0$ does not make sense, then $(a-b)^2=0$ so that $a=b$ which contradicts the hypotesis. Therefore $\sqrt{ab} < \frac{a+b}{2}$.

My only concern is whether it is correct to square an inequality in which both sides are positive.

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    $\begingroup$ I just discovered this is a duplicate question. $\endgroup$
    – ncmathsadist
    Commented May 1, 2022 at 0:26
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    $\begingroup$ $f(x)=x^2$ is an increasing function of $x$ when $x>0$, as is $g(x)=\sqrt{x}$ so yes you can square or square-root both sides of an inequality when both are non-negative. But you do not need to aim for a contradiction here. $(a-b)^2>0 \implies a^2+2ab+b^2>4ab$ will get you the result you want $\endgroup$
    – Henry
    Commented May 1, 2022 at 0:28

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Your proof is correct but you don't need to use contradiction. Instead, argue that

$$\sqrt{ab}< \sqrt{ab}+\left(\sqrt{\frac{a}{2}}-\sqrt{\frac{b}{2}}\right)^2=\frac{a+b}{2}$$

The strict inequality follows since $a\neq b$.

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You don't need to argue by contradiction. Expand this:
$$0\le \left(\sqrt{a} - \sqrt{b}\right)^2.$$ This will give you a direct proof.

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    $\begingroup$ Typo alert, think it should $\cdots \sqrt{a} - \sqrt{b} \cdots $ $\endgroup$
    – user2661923
    Commented May 1, 2022 at 0:30

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