Prove

\int_{x}^{1}{f(t)}dt \geq \frac{1-x^{2}}{2}

applying liebinz rule,-f(x)\geq -x

squaring both sidesf^{2}(x)\geq x^{2}

integratring between the limits 0 and 1 we get \int_{0}^{1}{f^{2}(x)}dx\geq \int_{0}^{1}{}x^{2}dx

therefore,\int_{0}^{1}{f^{2}(x)}dx\geq \frac{1}{3}

equality holds when f(x)=x

am i correct, sir??

hehe[1]

The correct answer does not always means that the method is correct. For example,
\dfrac{64}{16}=\dfrac{64\!\!\!\!\!\!/}{16\!\!\!/}=4
gives the correct result but obviously is wrong.

The point is that you cannot differentiate an inequality. Otherwise, we see that, for example,
\cos x \ge \sin x \quad \forall \ x\in [0,\,\pi/4]
Differentiating both sides w.r.t x, we get
-\sin x \ge \cos x \quad \forall \ x\in [0,\,\pi/4]
which is obviously wrong.

That's why both the solutions above are wrong.

this hint has the whole soln in itself

f(x)\geq g(x) for x\epsilon [a,b]

f(x+h)\geq g(x+h) for (x+h)\epsilon [a,b]

-g(x)\geq -f(x)

so, the inequality f(x+h)-f(x)\geq g(x+h)-g(x) is not correct..
.\frac{f(x+h)-f(x)}{h}\geq \frac{g(x+h)-g(x)}{h}

f`(x)\geq g`(x)
is not true

is this why we cannot differentiate an inequality??
is integration allowed on inequalities??

also pls post the right method for the problem

One can integrate both sides of an inequality if and only if the inequality is true for the entire interval of integration.
For the present problem, since for all x\in[0,\,1],
\int_x^1f(t)\ \mathrm{d}t \geq \dfrac{1-x^2}{2}
we get
\int_0^1\left(\int_x^1f(t)\ \mathrm{d}t\right)\mathrm{d}x \geq \int_0^1\dfrac{1-x^2}{2}\mathrm{d}x ------------- (1)
Integrate the LHS by parts taking 1 as the second function and \int_x^1f(t)\ \mathrm{d}t as the first one:
\left|x\int_x^1f(t)\ \mathrm{d}t\right|_0^1 -\int_0^1\left(\int_x^1f(t)\ \mathrm{d}t\right)^\prime x \ \mathrm{d}x
The first term evaluates to zero and noting that \left(\int_x^1f(t)\ \mathrm{d}t\right)^\prime =-f(x), we get from (1)
\int_0^1xf(x)\ \mathrm{d}x \geq \dfrac{1}{3}
Finally, from Cauchy Schwarz, we get
\dfrac{1}{9}\leq \left(\int_0^1f(x)\ \mathrm{d}x\right)^2\leq \int_0^1x^2\ \mathrm{d}x \int_0^1f^2(x)\ \mathrm{d}x
i.e.
\dfrac{1}{9}\leq \dfrac{1}{3} \int_0^1f^2(x)\ \mathrm{d}x
from where it follows that \int_0^1f^2(x)\ \mathrm{d}x \ge \dfrac{1}{3}