Pre-IMO 2017 Mock Exam

By the Cauchy-Schwarz inequality, we have $$(a^{2}b+c^{2}d+a{d}^2+c{b}^2)(a^{2}d+c^{2}b+a{b}^2+c{d}^2)\ge (a^2\sqrt{bd}+c^2\sqrt{bd}+abd+CBD{)}^2={\left(\frac{a^2+c^2}{\sqrt{ac}}+abd+CBD\right)}^2.$$ Together with $a^2+c^2\ge \frac{1}{2}(a+c{)}^2\ge (a+c)\sqrt{ac}$, we have $$(a^{2}b+b^{2}c+c^{2}d+d^{2}a)(a{b}^2+b{c}^2+c{d}^2+d{a}^2)\ge (a+c{)}^2(1+bd{)}^2.$$ Due to symmetry, we also have $$(a^{2}b+b^{2}c+c^{2}d+d^{2}a)(a{b}^2+b{c}^2+c{d}^2+d{a}^2)\ge (b+d{)}^2(1+ac{)}^2.$$ Multiplying these inequalities, we obtain $$(a^{2}b+b^{2}c+c^{2}d+d^{2}a)(a{b}^2+b{c}^2+c{d}^2+d{a}^2)\ge (a+c)(b+d)(1+ac)(1+bd)$$ where $(1+ac)(1+bd)=1+ac+bd+abcd=ac+bd+2$. The result follows readily. For equality in the application of the Cauchy-Schwarz inequality, we need $b=d$ and $a=c$. Note that the other inequalities also have these as equality. Together with $abcd=1$, equality holds when $a=c=t$ and $b=d=\frac{1}{t}$ for some $t>0$. $\square$