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This problem is from Antonio Caminha's book "Números reais" (real numbers), a Brazillian book for contest math preparation.

Problem:

If $a$,$b$,$c$ and $d$ are non-negative real numbers such that $a \le 1$, $a + b \le 5$, $a + b + c\le 14$ and $a + b + c +d\le 30$, prove that

$$\sqrt{a}+ \sqrt{b} + \sqrt{c} + \sqrt{d} \le 10 $$

In the problem, it suggests to use the Abel's inequality. But I couldn't find out the solution, I just noticed that if we take $a=1$, $b=4$, $c=9$ and $d=16$, we have $\sqrt{a}+ \sqrt{b} + \sqrt{c} + \sqrt{d} = 10 $.

The problem is already solved here: Inequality question­. However, there the problem is solved using advanced math (like polyhedral domain, stationary points, etc). Apparantly, this problem can be solved just with elementary math (things at the level of Abel's inequality, AM–GM inequality, etc), since the book is at high school level or even lower.

Rafael Deiga
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  • @MartinR there the problem is solved using quite advanced facts, i was expecting a more elementary solution. – Rafael Deiga Oct 20 '18 at 13:15
  • Not an answer to the question, but with Jensens inequality, because $f(x)=\sqrt{x}$ you would get that $$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leq 4\sqrt{\frac{a+b+c+d}{4}} \leq 4\sqrt{7.5} \approx 10.95$$ where I've only used the fact that $a+b+c+d\leq 30$. This is not that far away from 10. Maybe this argument can be tweaked a bit to get the desired upper bound? For example, may there be a way to "push" the $a+b+c+d \leq 30$-inequality even lower? If you can someshow that $a+b+c+d \leq 25$ then the inequality you wish to prove will follow. – Markus Oct 20 '18 at 13:23
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    This answer https://math.stackexchange.com/a/871986/42969 to the other question uses only elementary arguments. – Martin R Oct 20 '18 at 13:28
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    @RafaelDeiga I have posted a simple solution to the linked thread since this one is closed. Take a look at https://math.stackexchange.com/a/2963308/210525 – timon92 Oct 20 '18 at 14:11

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