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I want to prove that $\sqrt{a} + \sqrt{b} \le 2 \times \sqrt{a+b}$, I had the idea to draw it:
enter image description here

Would it be enough to prove what I want to prove? If not, is there a way to be more precise by still using my method or should I abandon it and use a more "traditional" way?

Thank you.

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  • $\begingroup$ There is no need for absolute values, as $a,b$ must both be nonnegative. $\endgroup$
    – vadim123
    Commented Feb 26, 2017 at 15:37
  • $\begingroup$ @vadim123 Thank you! I corrected the drawing. $\endgroup$ Commented Feb 26, 2017 at 15:41
  • $\begingroup$ have you thought about just squaring the inequality? $\endgroup$
    – user190080
    Commented Feb 26, 2017 at 15:42
  • $\begingroup$ @user190080 I did, and it probably is easier, but since my first idea was to draw it I am wondering if this sort of proof makes any sense... $\endgroup$ Commented Feb 26, 2017 at 15:43
  • $\begingroup$ I think this is a nice "proof without words," but in most contexts where you actually need to demonstrate this to someone you should use "traditional" methods. $\endgroup$
    – David K
    Commented Feb 26, 2017 at 16:45

2 Answers 2

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Note that $a,b$ are both nonnegative. $$\sqrt{a}\le \sqrt{a+b}$$ $$\sqrt{b}\le \sqrt{a+b}$$ Now add the two.

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  • $\begingroup$ Very elegant answer $\endgroup$
    – mrnovice
    Commented Feb 26, 2017 at 15:45
  • $\begingroup$ Simplest approach possible (+unity) :-) $\endgroup$
    – user399078
    Commented Feb 26, 2017 at 15:45
  • $\begingroup$ Nicely done. I personally would have squared both sides then use the AM-GM inequality (Which is much more complicated than this method). $\endgroup$ Commented Feb 26, 2017 at 15:48
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Not all problems have a nice geometrical interpretation. I think it is surely better to do this one in a purely algebraic way.

It would also be more convenient to assume a,b $\geq$ 0 for your problem.

Because a,b$\geq$0, then a$\leq$a+b and b$\leq$a+b, and because f(x)=$\sqrt{x}$ is increasing on $ R _ +$, then $\sqrt{a} \leq \sqrt{a+b}$ and $\sqrt{b} \leq \sqrt{a+b}.$ Finally, by adding the 2 inequalities, we get $\sqrt{a}+\sqrt{b} \leq 2*\sqrt{a+b}$

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