Question Number 140959 by loveineq last updated on 14/May/21
$$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}}{\mathrm{4}+{a}+{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by MJS_new last updated on 14/May/21
$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy} \\ $$$${a},\:{b}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{1}+{a}>\mathrm{0}\wedge\mathrm{1}+{b}>\mathrm{0}\wedge\mathrm{4}+{a}+{b}>\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{4}+{a}+{b}\right)\geqslant\mathrm{4}\left(\mathrm{4}−{a}−{b}\right)\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{5}{a}^{\mathrm{2}} {b}+\mathrm{6}{a}^{\mathrm{2}} +\mathrm{5}{ab}^{\mathrm{2}} +\mathrm{6}{b}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{true}\:\mathrm{since}\:{a},\:{b}\:\geqslant\mathrm{0} \\ $$