Square roots of variables?

Note that $f(x)=\sqrt{x}$ is a concave function, so by weighted Jensen's inequality, we have

$\dfrac {x_1 f(y_1) + x_2 f(y_2) + x_3 f(y_3) + x_4 f(y_4)}{x_1 + x_2 + x_3 + x_4} \leq f \left (\dfrac {x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4}{x_1 + x_2 + x_3 + x_4} \right )$

If we substitute $y_1=a$ , $y_2=\frac {b}{4}$ , $y_3 = \frac {c}{9}$ , $y_4 = \frac {d}{16}$ , $x_1 = \frac {1}{10}$ , $x_2 = \frac {2}{10}$ , $x_3 = \frac {3}{10}$ and $x_4 = \frac {4}{10}$ , we have

$\begin{aligned} \dfrac {1}{10}\sqrt{a} + \dfrac {2}{10} \sqrt{\dfrac {b}{4}} + \dfrac {3}{10} \sqrt{\dfrac {c}{9}} + \dfrac {4}{10} \sqrt{\dfrac {d}{16}} &\leq \sqrt{\dfrac {a}{10} + \dfrac {b}{20} + \dfrac {c}{30} + \dfrac {d}{40}}\\ \sqrt{a}+\sqrt{b}+ \sqrt{c} + \sqrt{d} &\leq 10 \sqrt{\dfrac {12a + 6b + 4c + 3d}{120}} \end{aligned}$

Note that $12a + 6b + 4c + 3d = 3(a+b+c+d)+(a+b+c)+2(a+b)+6a \leq 3 \times 30 + 14 + 2 \times 5 + 6 \times 1 = 120$ . Thus,

$\sqrt{\dfrac {12a + 6b + 4c + 3d}{120}} \leq 1$

so, $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq 10$ .

Just thought a=1,b=4,c=9,d=16..

Attempted 10 and was correct!

New solution:

We will prove a more general statement:

If ${ a }_{ 1 },{ a }_{ 2 },...,{ a }_{ n }$ are positive, $0\le { b }_{ 1 }\le { b }_{ 2 }\le ...\le { b }_{ n }$ , and for all $k \le n$ ,

${ a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ k }\le { b }_{ 1 }+{ b }_{ 2 }+...+{ b }_{ k }$ , then

$\large\ \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ 2 } } +...+\sqrt { { a }_{ n } } \le \sqrt { { b }_{ 1 } } +\sqrt { { b }_{ 2 } } +...\sqrt { { b }_{ n } }$

Let us prove the above result. We have

$\large\ \sum _{ r=1 }^{ n }{ \frac { { a }_{ r } }{ \sqrt { { b }_{ r } } } } =\quad { a }_{ 1 }\left( \frac { 1 }{ \sqrt { { b }_{ 1 } } } -\frac { 1 }{ \sqrt { { b }_{ 2 } } } \right) +\left( { a }_{ 1 }+{ a }_{ 2 } \right) \left( \frac { 1 }{ \sqrt { { b }_{ 2 } } } -\frac { 1 }{ \sqrt { { b }_{ 3 } } } \right) +....+\left( { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } \right) \left( \frac { 1 }{ \sqrt { { b }_{ n } } } \right)$ .

The differences in the parentheses are all positive. Using the hypothesis, we obtain that this expression is less than or equal to

$\large\ \quad { b }_{ 1 }\left( \frac { 1 }{ \sqrt { { b }_{ 1 } } } -\frac { 1 }{ \sqrt { { b }_{ 2 } } } \right) +\left( b_{ 1 }+{ b }_{ 2 } \right) \left( \frac { 1 }{ \sqrt { { b }_{ 2 } } } -\frac { 1 }{ \sqrt { { b }_{ 3 } } } \right) +....+\left( { b }_{ 1 }+{ b }_{ 2 }+...+b_{ n } \right) \left( \frac { 1 }{ \sqrt { { b }_{ n } } } \right) =\sqrt { { b }_{ 1 } } +\sqrt { { b }_{ 2 } } +...+\sqrt { { b }_{ n } }$ .

Hence,

$\large\ \sum _{ r=1 }^{ n }{ \frac { { a }_{ r } }{ \sqrt { { b }_{ r } } } } \le \sqrt { { b }_{ 1 } } +\sqrt { { b }_{ 2 } } +...\sqrt { { b }_{ n } }$ .

Using this result and and the CAUCHY-SCHWARZ inequality, we obtain

$\large\ { \left( \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ 2 } } +...+\sqrt { { a }_{ n } } \right) }^{ 2 }={ \left( \sum _{ r=1 }^{ n }{ \left( \sqrt [ 4 ]{ { b }_{ r } } .\sqrt { \frac { { a }_{ r } }{ \sqrt { { b }_{ r } } } } \right) } \right) }^{ 2 }$

$\large\ \le$

$\large\ \left( \sqrt { { b }_{ 1 } } +\sqrt { { b }_{ { 2 } } } +...+\sqrt { { b }_{ n } } \right) \left( \sum _{ r=1 }^{ n }{ \left( \frac { { a }_{ r } }{ \sqrt { { b }_{ r } } } \right) } \right) \le { \left( \sqrt { b_{ 1 } } +\sqrt { { b }_{ 2 } } +...+\sqrt { b_{ n } } \right) }^{ 2 }$ .

This gives

$\large\ \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ 2 } } +...+\sqrt { { a }_{ n } } \le \sqrt { { b }_{ 1 } } +\sqrt { { b }_{ 2 } } +...\sqrt { { b }_{ n } }$ .

The special case of the original problem is obtained for $n = 4$ , by setting ${ b }_{ k } = { { k }^{ 2 } }$ , $k = 1, 2, 3, 4$

Which gives us $\large\ \sqrt { a } + \sqrt { b } + \sqrt { c } + \sqrt { d } \le 10$ .

Please tell how the solution is.