<s>[INST] 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.
 [/INST] 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*}
</s>