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Just out of curiosity, does the following identity hold?$$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},\quad a \gt 0$$

Thanks

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6 Answers 6

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$$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = \frac{2}{\sqrt{2}}\sqrt{a}$$

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    $\begingroup$ $=\sqrt{2}\sqrt{a}$ $\endgroup$
    – kram1032
    Mar 22, 2013 at 23:14
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$$\sqrt{\frac{a}{2}} + \sqrt{\frac a2} = 2 \frac{\sqrt a}{\sqrt 2} = \frac {2}{\sqrt 2} \sqrt{a} = \sqrt 2 \sqrt a = \sqrt{2a} > \sqrt a$$


Added: perhaps your question originates from not remembering that $$(a + b)^2 \not\equiv a^2 + b^2\;?$$ If we do square each side of the proposed inequality (which we can do, without having to worry about whether doing so might change the direction of the inequality, since $a\gt 0$), then note: $$\left(\sqrt{\frac a2} + \sqrt{\frac a2}\right)^2\;\; \geq \; \;(\sqrt a)^2$$ $$ \iff \frac a2 + 2\cdot \frac a2 + \frac a2 \;\;\geq \;\; a $$ $$\iff 2a \geq a\quad \checkmark$$

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    $\begingroup$ Interesting note though: $$\sqrt{\dfrac{a}{4}} + \sqrt{\dfrac{ a}{4}} = \sqrt a$$ $\endgroup$
    – amWhy
    Mar 22, 2013 at 18:24
  • $\begingroup$ Or you could just end with $\sqrt{2}\sqrt{a}=\sqrt{2a} > \sqrt{a}$ instead of having to explain that $\sqrt{2}>1$. $\endgroup$
    – chharvey
    Mar 23, 2013 at 2:30
  • $\begingroup$ @amWhy It does, thanks $\endgroup$
    – bobdylan
    Mar 24, 2013 at 18:36
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Yes, you can distribute out $\sqrt a(\sqrt {\frac 12}+\sqrt {\frac 12})=\sqrt 2 \sqrt a \gt \sqrt a$

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$$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = \frac{2}{\sqrt{2}}\sqrt{a}\geq \sqrt{a}$$

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Simple answer: yes. See this for explanation:

$$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge\sqrt{a} \; \wedge \; a>0$$ $$\Leftrightarrow \; 2\frac{\sqrt{a}}{\sqrt{2}}\ge\sqrt{a} \; \wedge \; a>0$$ $$\Leftrightarrow \; 2\frac{\sqrt{2}\sqrt{a}}{2}\ge\sqrt{a} \; \wedge \; a>0$$ $$\Leftrightarrow \; \sqrt{2}\sqrt{a}\ge\sqrt{a} \; \wedge \; a>0$$ $$\Leftrightarrow \; \sqrt{2}\ge1 \; \wedge \; a>0$$ Therefore, true.

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    $\begingroup$ This is good informal reasoning but not a good proof. If you start out by assuming a conjecture is true and then arrive at a tautology, that does not necessarily prove that the conjecture is true. It just shows that the consequence of it being true is a tautology. $\endgroup$
    – chharvey
    Mar 23, 2013 at 2:28
  • $\begingroup$ @TestSubject528491 It this better then? Inserter bi-implications and the condition that $a>0$ in every step. $\endgroup$
    – tomsmeding
    Mar 23, 2013 at 12:40
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Take $a=8b^2,$ where $b >0.$
Now $$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = 2\sqrt\frac {8b^2}{2}=4b > 2\sqrt 2 b=\sqrt a.$$

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