# Inequalities – Property IV

## Inequalities – Property IV

Property IV: If $${{f}^{2}}\left( x \right)$$ and $${{g}^{2}}\left( x \right)$$ are integrable on the interval [a, b]. then $$\left| \int\limits_{a}^{b}{f\left( x \right)g\left( x \right).dx} \right|\le \sqrt{\left( \int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx} \right)\left( \int\limits_{a}^{b}{{{g}^{2}}\left( x \right).dx} \right)}$$.

Proof:

Let $$F\left( x \right)={{\left( f\left( x \right)-\lambda g\left( x \right) \right)}^{2}}\ge 0$$,

Where $$\lambda$$, is real number. Then

$$\int\limits_{a}^{b}{{{\left( f\left( x \right)-\lambda g\left( x \right) \right)}^{2}}}.dx\ge 0$$,

$$\int\limits_{a}^{b}{{{f}^{2}}\left( x \right)-2\times f\left( x \right)\times \lambda g\left( x \right)+{{\lambda }^{2}}{{g}^{2}}\left( x \right)}.dx\ge 0$$,

$$\int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx-2\lambda \int\limits_{a}^{b}{{}}f\left( x \right)g\left( x \right).dx+{{\lambda }^{2}}\int\limits_{a}^{b}{{}}{{g}^{2}}\left( x \right)}.dx\ge 0$$,

Therefore, discriminant is non positive, i.e., $${{B}^{2}}-4AC\le 0$$,

$$4{{\left( \int\limits_{a}^{b}{f\left( x \right)g\left( x \right).dx} \right)}^{2}}\le 4\int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx\int\limits_{a}^{b}{{{g}^{2}}\left( x \right)}.dx}$$,

Hence

$$\left| \int\limits_{a}^{b}{f\left( x \right)g\left( x \right).dx} \right|\le \sqrt{\left( \int\limits_{a}^{b}{{{f}^{2}}\left( x \right).dx} \right)\left( \int\limits_{a}^{b}{{{g}^{2}}\left( x \right).dx} \right)}$$.

Example: find the value of  $$\int\limits_{0}^{1}{\sqrt{\left( 1+x \right)\left( 1+{{x}^{3}} \right)}.dx}$$,

Solution:

$$\int\limits_{0}^{1}{\sqrt{\left( 1+x \right)\left( 1+{{x}^{3}} \right)}.dx}$$,

$$\int\limits_{0}^{1}{\sqrt{\left( 1+x \right)\left( 1+{{x}^{3}} \right)}.dx}\le \sqrt{\left( \int\limits_{0}^{1}{\left( x+1 \right)dx} \right)\left( \int\limits_{0}^{1}{\left( \ \ \ \ \ \ \ \ 1+{{x}^{3}} \right)dx} \right)}$$,

$$=\sqrt{{{\left( x+\frac{{{x}^{2}}}{2} \right)}_{0}}^{1}{{\left( x+\frac{{{x}^{4}}}{4} \right)}_{0}}^{1}}$$,

$$=\sqrt{\left( 1+\frac{{{1}^{2}}}{2} \right)\left( 1+\frac{{{1}^{4}}}{4} \right)}$$,

$$=\sqrt{\frac{15}{8}}$$.