Let A be the set of twice continuously differentiable functions on the interval $[0, 1]$,  and let $B = \{f\in A : f (0) = f (1) = 0, f ′ (0) = 2\}$. What is  min $\int_0^1 (f''(x))^2$.   There is one given hint consider $(1-x)f''(x)$.  Any idea how to start solving this problem?
I was trying to solve a question of an entrance exam. I am completely stuck in the problem. I am not able to find idea how to proceed. Please help me.     Let $A$ be the set of twice continuously differentiable functions on the interval $[0, 1]$ and let $B = \{f \in A : f(0) = f(1) = 0, f'(0) = 2\}$. Find the value of $$\text{min}_{f\in B} \displaystyle \int_0^1 (f''(x))^2dx.$$    I am really sorry for not showing some effort from my side but I can not find ant way to proceed.  Please help me. Thnx in advance.
Number of possible combinations of x numbers that sum to y I want to find out the number of possible combinations of $x$ numbers that sum to $y$. For example, I want to calculate all combination of 5 numbers, which their sum equals to 10.  An asymptotic approixmation is also useful. This question seems to be very close to number partitioning, with the difference that a number can be 0. See:    All possible partitions for sum 10 and 3 positions that can be zero, are 63 possiblities: (numbers shown as 3 digits)  019 028 037 046 055 064 073 082 091 109 118 127 136 145 154 163 172 181 190 208 217 226 235 244 253 262 271 280 307 316 325 334 343 352 361 370 406 415 424 433 442 451 460 505 514 523 532 541 550 604 613 622 631 640 703 712 721 730 802 811 820 901 910 $ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $, prove $ab + bc + ca \le 3$ If $a,b,c$ real numbers such that $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$,  prove that $ab + bc + ca \le 3$.     Attempt:  Assume that $a,b,c &gt; 0$.   $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1  $$ $$ \implies  (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1) + (a^{2} + b^{2} + 1)(a^{2}+c^{2}+1) +  (a^{2} + c^{2} + 1)(b^{2}+c^{2}+1) \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  $$ \implies  ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1) + ((ab)^{2} + (ac)^{2} + (bc)^{2} + b^{2} + 2 a^{2} + c^{2} + a^{4} + 1) + ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 c^{2} + b^{2} + c^{4} + 1) \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  $$ \implies  3((ab)^{2} + (ac)^{2} + (bc)^{2}) + 4(a^{2} + b^{2} + c^{2}) + (a^{4}+b^{4}+c^{4}) + 3  \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  Next, notice  $$(a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1) = ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1)(c^{2} + a^{2} + 1) $$ $$ = (abc)^{2} + a^{2}c^{4} + b^{2}c^{4} + (ac)^{2} + 2(bc)^{2} + c^{4} + b^{4}c^{2} + c^{2} + a^{4} b^{2} + a^{4}c^{2} + (abc)^{2} + a^{4}  + 2(ab)^{2} + (ac)^{2} + b^{4}a^{2} + a^{2} +  (ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1 $$ $$ = 2(abc)^{2} + 3(ac)^{2} + 3(bc)^{2} + 3(ab)^{2} + c^{4}a^{2} + c^{4} b^{2} + b^{4}a^{2} + b^{4} c^{2} + a^{4}b^{2} + a^{4} c^{2} + a^{4} + b^{4} + c^{4} + 2(a^{2} + b^{2} + c^{2}) + 1  $$  So we have  $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$ $$ \implies  2(a^{2} + b^{2} + c^{2}) +  3  \ge 2(abc)^{2}  + c^{4}a^{2} + c^{4} b^{2} + b^{4}a^{2} + b^{4} c^{2} + a^{4}b^{2} + a^{4} c^{2} + 1 $$ Answer draft is lost if you click to sign up How can I make a PATCH HTTP Callout from Apex? 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