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How do we prove that $$a+b+\frac12 \geq \sqrt{a} + \sqrt{b}$$ with $a\geq 0$ and $b\geq 0$?

I tried using Cauchy inequality but doesn't seem to work. Factor things out also seem impossible or require complex numbers.

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    $\begingroup$ Call $x=\sqrt{a}, y=\sqrt{b}$, the inequality is equivalent to $x^2+y^2+1/2\ge x+y$, which is equivalent to $(x-1/2)^2+(y-1/2)^2\ge 0$. $\endgroup$
    – Q-Zhang
    Commented Sep 21, 2021 at 2:10
  • $\begingroup$ Alternative approach - not as elegant as the approach of @Q.Zhang : when $r > 1,$ you have that $r > \sqrt{r}$. Therefore, without loss of generality, $1 > a,b.$ For any $r$ in the interval $(0,1)$, you have that $r - r^2 = r(1-r).$ So, you have two numbers, $r$ and $(1 - r)$, whose arithmetic mean is $(1/2).$ Based on the idea, that in general $(r + k)(r - k) = r^2 - k^2 < r^2$, you have that the arithmetic mean of two positive numbers is always greater than or equal to the geometric mean. Therefore, the geometric mean of $(r), (1-r)$ must be $\leq (1/2)$. ...see next comment $\endgroup$ Commented Sep 21, 2021 at 2:29
  • $\begingroup$ This implies that the maximum value that can be obtained by $\sqrt{a} - a = \sqrt{a} - \left(\sqrt{a}\right)^2$ is $(1/4).$ Ditto for $\sqrt{b} - b.$ $\endgroup$ Commented Sep 21, 2021 at 2:33
  • $\begingroup$ Please add your exact effort to your question. Then, your question will seem very nice. $\endgroup$ Commented Sep 21, 2021 at 3:10

1 Answer 1

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Let $a,b\geq 0$, then $\sqrt{a},\sqrt{b}\in\mathbb{R}$ and therefore, $$\left(\sqrt{a}-\frac12\right)^2+\left(\sqrt{b}-\frac12\right)^2\geq 0$$ $$\implies a+b-(\sqrt{a}+\sqrt{b})+\frac12\geq 0$$ $$\implies a+b+\frac12\geq \sqrt{a}+\sqrt{b}$$

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