Let $f:[0,1]\to R$ be a continuous and differentiable function such that for any $x,y \in [0,1], xf(y)+yf(x)\le 1$, then prove the following.... 
Let $f:[0,1]\to R$ be a continuous and differentiable function such that for any $x,y \in [0,1], xf(y)+yf(x)\le 1$, then prove the following


1)$\int _0^1 f(x) dx \le \pi/4$


2)$\int_0^1 f'(x) x^2 dx \ge f(1)-1$


3)$\int _0^1 x f'(x) dx\ge f(1)-\pi/4$

Proving any one of them will be fine, I just need a base to solve. I did solve the second one using integration by part and by deriving the relation $f(x)\le \frac{1}{2x}$, but i am having trouble with the rest
Do we need to assume a function and solve or can the function $f(x)$ be derived from the given data? It looks like a lot of guesswork but I would like to know if there is a solid proof.

Although the first has been answered in $xf(y)+yf(x)\leq 1$ for all $x,y\in[0,1]$ implies $\int_0^1 f(x) \,dx\leq\frac{\pi}{4}$ in a very simple way, an alternative way (which is slightly complicated) exists too. The reason for writing this answer is just to share how thoughts ran in my mind. I took $\frac{\pi}{4}$ as my first hint and thought that some sort of $\frac{1}{1+x^2}$ should come on the right hand side. For this I just started choosing values for  points $x$ and $y$ in terms of $x$. After certain trials, some thoughtful process led me choose the following:
Both $\frac{2x}{1+x^2}$ and $\frac{1-x^2}{1+x^2}$ lie in $[0,1]$. Using the given hypothesis, we can write:
$$\frac{2x}{1+x^2} f\left(\frac{1-x^2}{1+x^2}\right) + \frac{1-x^2}{1+x^2}f\left(\frac{2x}{1+x^2}\right) \leq 1.$$
Or said equivalently,
$$\frac{2x}{(1+x^2)^2} f\left(\frac{1-x^2}{1+x^2}\right) + \frac{1-x^2}{(1+x^2)^2}f\left(\frac{2x}{1+x^2}\right) \leq \frac{1}{1+x^2}.$$
If we now notice that,
$$ \int\limits_{0}^{1} \frac{2x}{(1+x^2)^2} f\left(\frac{1-x^2}{1+x^2}\right)dx = \int\limits_{0}^{1} \frac{1-x^2}{(1+x^2)^2}f\left(\frac{2x}{1+x^2}\right) dx = \frac{1}{2}\int\limits_{0}^{1} f(t) dt$$,
we come to the conclusion that
$$\int\limits_{0}^{1}f(x)dx \leq \int\limits_{0}^{1} \frac{1}{1+x^2}dx = \tan^{-1}{1}-\tan^{-1}0 = \frac{\pi}{4}$$.
As has been pointed out rightly, third one follows easily.
