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.
 A: The function $f(a,b,c,d)=\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}$ has no stationary points inside the polyhedral domain (since the partial derivatives cannot vanish due to the concavity of the square root function), hence its maximum is attained on the boundary. By iterating the same argument on the boundary of the domain, we have that the maximum is achieved in a vertex of the polyhedron, and by checking them all we have that $$\operatorname{argmax}(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})=(1,4,9,16),$$ from which
$$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq 1+2+3+4 = 10$$
follows.
A: Using AM-QM and the assumptions we get
\begin{align*}
\sqrt a + \sqrt b + \sqrt c + \sqrt d & 
= \sqrt a + 2\sqrt{\frac b4} + 3\sqrt{\frac c9} + 4 \sqrt{\frac d{16}} \\
& \le 10 \sqrt{\frac{a+2\cdot \frac b4 + 3 \cdot \frac c9 + 4 \cdot \frac d{16}}{10}} \\
& = \sqrt{10a+5b+\frac{10}3 c + \frac 52 d} \\
& = \sqrt{5a+\frac 53 (a+b) + \frac 56 (a+b+c) + \frac 52(a+b+c+d)}  \\
& \le \sqrt{5 \cdot 1 + \frac 53 \cdot 5 + \frac 56 \cdot 14 + \frac 52 \cdot 30}\\
& = 10.
\end{align*}
A: 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 $a_0 = 1$.
A: $d$ should be maximized, so the last inequality becomes an equation.
Keep $a,b$ fixed, and vary $c$ and $d=30-a-b-c$, perhaps $c$ must be maximized so the third inequality becomes an equation.
A: a<=1 ,a+b<=5 
Or, b<=4
a+b<=5 , a+b+c<=14
c<=9
a+b+c<=24 ,a+b+c+d<=30
d<=16
sqrt a +sqrt b +sqrt c+sqrt d<=sqrt 1+sqrt 4+sqrt 9+sqrt 16
sqrt a+sqrt b+sqrt c+sqrt d <=10 
