# Contradiction Observation to Cauchy Schwarz inequality for inequalty

$$f:[0,1]\to \mathbb R$$ , 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

Any help will be appreciated

• It's not? Just let $g=1$ for your result. – user608030 Dec 6 '18 at 15:19

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.
• Apply the Cauchy-Schwarz inequality to the functions $\lvert f\rvert$ and $1$. – José Carlos Santos Dec 6 '18 at 15:45