$f:[0,1]\to \mathbb R$ , be continuous function then prove that $$\int_0^1f^2(x)dx\geq \biggl(\int_0^1|f(x)| \biggr) ^2$$
I tried this for $x^2$
For that above is true
But I checked following proof Which is complete opposite to above. Proving the Cauchy-Schwarz integral inequality in a different way
Please help me to find that where is I am making wrong ?
Any help will be appreciated
The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,g\in C^1\bigl([0,1]\bigr)$, then$$\left(\int_0^1f(x)g(x)\,\mathrm dx\right)^2\leqslant\left(\int_0^1f^2(x)\,\mathrm dx\right)\left(\int_0^1g^2(x)\,\mathrm dx\right).$$If you choose $g=f$, you get the trivial inequality$$\left(\int_0^1f^2(x)\,\mathrm dx\right)^2\leqslant\left(\int_0^1f^2(x)\,\mathrm dx\right)^2$$or$$\int_0^1f^2(x)\,\mathrm dx\leqslant\int_0^1f^2(x)\,\mathrm dx.$$This in no way contradicts what you are supposed to prove.