Inequality question­ $$a,b,c,d\ge 0$$
$$a\le 1$$
$$a+b\le 5$$
$$a+b+c\le 14$$
$$a+b+c+d\le 30$$
Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10$.
We can subtract inequalities to get the answer, but that is wrong... I can't think of any another method... Any hints or suggestions will be helpful.

Let $(a_0,b_0,c_0,d_0)$ be a point in the domain where $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is maximized. It suffices to show that $a_0 = 1$, $a_0 + b_0 = 5$, $a_0 + b_0 + c_0 = 14$, and $a_0 + b_0 + c_0 + d_0 = 30$ for then you can solve for all four variables to get $(a_0,b_0,c_0,d_0) = (1,4,9,16)$.
We start with the last equation and proceed backwards. Note that if $a_0 + b_0 + c_0 + d_0$ were strictly less than  $30$ we could increase $d_0$ slightly keeping the other variables fixed, and we'd stay in the domain yet $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ would increase, contradicting maximality of $\sqrt{a_0} + \sqrt{b_0} + \sqrt{c_0} + \sqrt{d_0}$. So we must have $a_0 + b_0 + c_0 + d_0 = 30$. Note that since $a_0 + b_0 + c_0 \leq 14$ this means $d_0 \geq 16$. In particular $d_0 > c_0$ since $c_0 \leq 14$.
Moving to the equation $a_0 + b_0 + c_0 \leq 14$, if strict inequality held here then one could replace $c_0$ by $c_0 + \epsilon$ and $d_0$ by $d_0 - \epsilon$ for some small $\epsilon$ and we would stay in the domain. However we must have $\sqrt{c_0 + \epsilon} + \sqrt{d_0 - \epsilon} > \sqrt{c_0} + \sqrt{d_0}$. To see why, squaring this inequality we see it's equivalent to
$$c_0 + d_0 + 2\sqrt{(c_0 + \epsilon)(d_0 - \epsilon)} > c_0 + d_0 + 2\sqrt{c_0d_0}$$
Subtracting $c_0 + d_0$ from both sides and squaring the result, we see this is the same as
$$(c_0 + \epsilon)(d_0 - \epsilon) > c_0d_0$$
Equivalently,
$$(d_0 - c_0)\epsilon - \epsilon^2 > 0$$
Since $d_0 \geq 16 > 14 \geq c_0$, this will hold if $\epsilon$ is small enough.
In summary, if we had $a_0 + b_0 + c_0 < 14$, then adjusting $c_0$ and $d_0$ as above would result in a new point $(a,b,c,d)$ in the domain for which $\sqrt{c} + \sqrt{d}$ is strictly larger than $\sqrt{c_0} + \sqrt{d_0}$. Since $a = a_0$ and $b = b_0$ are unchanged, $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is larger than $\sqrt{a_0} + \sqrt{b_0} + \sqrt{c_0} + \sqrt{d_0}$, contradicting the maximality of $\sqrt{a_0} + \sqrt{b_0} + \sqrt{c_0} + \sqrt{d_0}$.
We conclude $a_0 + b_0 + c_0 = 14$. To get the other two inequalities, you just iterate the above procedure to get $a_0 + b_0 = 5$, and then $a_0 = 1$.
