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 $$${x}>{0}$$$.