$ a^2\le 1 \Rightarrow a\le 1$ $ a^2\le 1, a^2 + b^2\le 5 \Rightarrow b^2\le 4 \Rightarrow b\le2$ $ a^2 + b^2\le 5, a^2 + b^2 + c^2\le 14 \Rightarrow c^2\le 9 \Rightarrow c\le3$ $ a^2 + b^2 + c^2\le 14, a^2 + b^2 + c^2 + d^2\le 30 \Rightarrow d^2\le 16 \Rightarrow d\le4$ $ \Rightarrow a + b + c + d\le 10$

mszew wrote: $ a^2\le 1, a^2 + b^2\le 5 \Rightarrow b^2\le 4$ I think, here you are wrong.

Sorry for my mistake...

$ (a-1)^2 + (b-2)^ + (c-3)^2 + (d-4)^2 + 2a + 4b + 6c + 8d - 30 \leq 30$ so $ a + 2b + 3c + 4d \leq 30$ The same way we get $ a + 2b + 3c \leq 14$ and $ a + 2b \leq 5$. There is also $ a\leq 1$. Now we have: $ 12(a + b + c + d) = 3(a + 2b + 3c + 4d) + (a + 2b + 3c) + 2(a + 2b) + 6a \leq 3*30 + 14 + 2*5 + 6 = 120$ So $ a + b + c +d \leq 10$

note that $ 12a^2 + 6b^2 + 4c^2 + 3d^2 = 120$ and that $ \frac {1}{12} + \frac {1}{6} + \frac {1}{4} + \frac {1}{3} = \frac {5}{6}$. from cauchy-schwarz we find that $ 100 = (12a^2 + 6b^2 + 4c^2 + 3d^2)\left(\frac {1}{12} + \frac {1}{6} + \frac {1}{4} + \frac {1}{3}\right)\geq (a + b + c + d)^2$ from where we conclude that $ a + b + c + d\leq |a + b + c + d|\leq 10$. edit: sorry, i didn't realize that this was rem solution at the above link.

campos wrote: note that $ 12a^2 + 6b^2 + 4c^2 + 3d^2 = 120$ and that $ \frac {1}{12} + \frac {1}{6} + \frac {1}{4} + \frac {1}{3} = \frac {5}{6}$. from cauchy-schwarz we find that $ 100 = (12a^2 + 6b^2 + 4c^2 + 3d^2)\left(\frac {1}{12} + \frac {1}{6} + \frac {1}{4} + \frac {1}{3}\right)\geq (a + b + c + d)^2$ from where we conclude that $ a + b + c + d\leq |a + b + c + d|\leq 10$. edit: sorry, i didn't realize that this was rem solution at the above link. I think it must be $ 12a^2 + 6b^2 + 4c^2 + 3d^2 \le 120$. Nice, campos. How about this (for the same conditions) : $ a+b+c+d+\frac{1}{10}((a+d-5)^2+(b+c-5)^2) \le 10$ There is a nice general problem with a very simple solution!!!

Let $ a,b,c$ be real numbers, such that $ a\ge b+1, b\ge c+4,$. Prove that \[a^2 + b^2 + c^2 \ge 14.\]

sqing wrote: Let $ a,b,c$ be real numbers, such that $ a\ge b+1, b\ge c+4,$. Prove that \[a^2 + b^2 + c^2 \ge 14.\] When $c< -4$ , $a^2 + b^2 + c^2 \ge c^2>14$; When $c\ge -4$, $a^2 + b^2 + c^2 \ge (c+5)^2+(c+4)^2+c^2= 3(c+3)^2+14\geq14 $. here

rem wrote: Let $ a,b,c,d$ be real numbers, such that $ a^2\le 1, a^2 + b^2\le 5, a^2 + b^2 + c^2\le 14, a^2 + b^2 + c^2 + d^2\le 30$. Prove that $ a + b + c + d\le 10$. We have, $$ 1 \geq a^2 \quad \quad \quad \quad - - - (1)$$ $$5\geq a^2+b^2 \quad \quad \quad \quad - - -(2)$$ $$14 \geq a^2+b^2+c^2 \quad \quad \quad - - - (3)$$ $$30\geq a^2+b^2+c^2+d^2 \quad \quad - - - (4)$$ From , $6 \times (1) + 2 \times (2) + (3) + 3 \times (4) $, we obtain $$120\geq 12a^2+6b^2+4c^2+3d^2 \quad \quad \quad \quad - - -(5)$$ Using Cauchy-Schwarz inequality, we have $$ (12a^2+6b^2+4c^2+3d^2)\left(\frac{1}{12} +\frac{1}{6} +\frac{1}{4}+\frac{1}{3}\right) \geq (a+b+c+d)^2$$ $$ \implies \frac{10}{12} (12a^2+6b^2+4c^2+3d^2) \geq (a+b+c+d)^2 \quad \quad \quad \quad - - - (6)$$ Combining (5) and (6), $$100 =\frac{10}{12}\cdot 120\geq \frac{10}{12} (12a^2+6b^2+4c^2+3d^2) \geq (a+b+c+d)^2$$ $$\implies 10 \geq a+b+c+d$$ proves the desired inequality.

rem wrote: Let $ a,b,c,d$ be real numbers, such that $ a^2\le 1, a^2 + b^2\le 5, a^2 + b^2 + c^2\le 14, a^2 + b^2 + c^2 + d^2\le 30$. Prove that $$ a + b + c + d\le 10$$ Let $a, \, b, \, c, \, d$ be real numbers, such that $a^2 \leq1,\,a^2+b^2 \leq 5,\,a^2+b^2+c^2\leq 14,\,a^2+b^2+c^2+d^2\leq 30.$ Prove that \[a+b+c+d+\frac{1}{10}\left[(a+d-5)^2+(b+c-5)^2\right] \leqslant 10.\]

sqing wrote: Let $a, \, b, \, c, \, d$ be real numbers, such that $a^2 \leq1,\,a^2+b^2 \leq 5,\,a^2+b^2+c^2\leq 14,\,a^2+b^2+c^2+d^2\leq 30.$ Prove that \[a+b+c+d+\frac{1}{10}\left[(a+d-5)^2+(b+c-5)^2\right] \leqslant 10.\] And there is a very simple proof：$$\Longleftrightarrow a^2+b^2+c^2+d^2+2ad+2bc\le 50$$$$2ad\le\frac{48a^2+3d^2}{12},2bc\le \frac{18b^2+8c^2}{12} $$$$ a^2+b^2+c^2+d^2+2ad+2bc \le \frac{60a^2+30b^2+20c^2+15d^2}{12}$$$$=\frac{15(a^2+b^2+c^2+d^2)+5(a^2+b^2+c^2)+10(b^2+c^2)+30a^2}{12}\le \frac{450+70+50+30}{12}=50$$