Proving Inequalities Using Monotony of the Function

Example 1. Prove that ${\sin{{\left({x}\right)}}}>\frac{{2}}{\pi}{x}$ when ${0}<{x}<\frac{\pi}{{2}}$.

Consider function ${f{{\left({x}\right)}}}=\frac{{{\sin{{\left({x}\right)}}}}}{{x}}$ defined on interval ${\left({0},\frac{\pi}{{2}}\right]}$.

We have that ${f{'}}{\left({x}\right)}=\frac{{{x}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{x}}^{{2}}}=\frac{{{\cos{{\left({x}\right)}}}{\left({x}-{\tan{{\left({x}\right)}}}\right)}}}{{{x}}^{{2}}}$. On interval ${\left({0},\frac{\pi}{{2}}\right)}$ ${\cos{{\left({x}\right)}}}>{0}$ and ${x}<{\tan{{\left({x}\right)}}}$, so ${f{'}}{\left({x}\right)}<{0}$. This means that function is decreasing on interval ${\left({0},\frac{\pi}{{2}}\right)}$.

So, for any ${x}$ from ${\left({0},\frac{\pi}{{2}}\right)}$ we have that ${f{{\left({x}\right)}}}=\frac{{{\sin{{\left({x}\right)}}}}}{{x}}>{f{{\left(\frac{\pi}{{2}}\right)}}}=\frac{{2}}{\pi}$ or equivalently ${\sin{{\left({x}\right)}}}>\frac{{2}}{\pi}{x}$.

Example 2. Prove that ${\cos{{\left({x}\right)}}}>{1}-\frac{{1}}{{2}}{{x}}^{{2}}$ for ${x}>{0}$.

Consider function ${f{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}-{1}+\frac{{1}}{{2}}{{x}}^{{2}}$. We have that ${f{{\left({0}\right)}}}={\cos{{\left({0}\right)}}}-{1}+\frac{{1}}{{2}}{{0}}^{{2}}={0}$.

derivative of function ${f{'}}{\left({x}\right)}=-{\sin{{\left({x}\right)}}}+{x}>{0}$ because ${\sin{{\left({x}\right)}}}<{x}$.

This means that for ${x}\ge{0}$ function is increasing and for ${x}>{0}$ we have that ${f{{\left({x}\right)}}}>{f{{\left({0}\right)}}}={0}$, i.e. ${\cos{{\left({x}\right)}}}-{1}+\frac{{1}}{{2}}{{x}}^{{2}}>{0}$. This is equivalent to ${\cos{{\left({x}\right)}}}>{1}-\frac{{1}}{{2}}{{x}}^{{2}}$.

Example 3. Prove that ${\ln{{\left({x}\right)}}}\le{x}-{1}$ for ${x}>{0}$.

Consider function ${f{{\left({x}\right)}}}={\ln{{\left({x}\right)}}}-{x}$. Its derivative is ${f{'}}{\left({x}\right)}=\frac{{1}}{{x}}-{1}$. Clearly ${f{'}}{\left({x}\right)}>{0}$ when ${0}<{x}<{1}$ and ${f{'}}{\left({x}\right)}<{0}$ when ${x}>{1}$.

So, function is increasing on interval ${\left({0},{1}\right)}$ and decreasing on interval ${\left({1},\infty\right)}$. So, at point ${x}={1}$ it takes its largest value ${f{{\left({1}\right)}}}={\ln{{\left({1}\right)}}}-{1}=-{1}$.

That's why ${\ln{{\left({x}\right)}}}-{x}\le-{1}$ for ${x}>{0}$. This is equivalent to ${\ln{{\left({x}\right)}}}\le{x}-{1}$ for ${x}>{0}$.