Hard Inequality

Question

Let $a,b,c,d$ be positive Real numbers, such that $ab+bc+cd+da+ac+bd=6abcd$.

Prove that $a+b+c+d-2\sqrt{2}abcd \leqslant 4-2\sqrt{2}$

Answer 1

We'll replace $a$ with $\frac{1}{a}$, $b$ with $\frac{1}{b}$, $c$ with $\frac{1}{c}$ and $d$ with $\frac{1}{d}$.

Hence, the condition gives $ab+ac+bc+ad+bd+cd=6$ and we need to prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{2\sqrt2}{abcd}\leq4-2\sqrt2$$ or $$(4-2\sqrt2)abcd+2\sqrt2\geq abc+abd+acd+bcd.$$ We'll prove that this inequality is true even for all non-negatives $a$, $b$, $c$ and $d$ such that

$ab+ac+bc+ad+bd+cd=6$.

Indeed, let $$f(a,b,c,d,\lambda)=(4-2\sqrt2)abcd+2\sqrt2-abc-abd-acd-bcd+$$ $$+\lambda(ab+ac+ad+bc+bd+cd-6).$$ But $f$ is a continuous function and $$A=\left\{(a,b,c,d,\lambda)\big|a\geq0,b\geq0,c\geq0,d\geq0,\sum\limits_{sym}ab=24\right\}$$ is a compact.

Thus, $f$ gets on $A$ the minimal value in some point $(a,b,c,d,\lambda)$,

which happens in the following cases.

1. $abcd=0$.

Let $d=0$.

Hence, by AM-GM $$abc=\sqrt{a^2b^2c^2}=\sqrt{ab\cdot ac\cdot bc}\leq\sqrt{\left(\frac{ab+ac+bc}{3}\right)^3}=\sqrt8=2\sqrt2$$

2. $abcd\neq0$ and $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial d}=\frac{\partial f}{\partial \lambda}=0,$$ which gives $$(4-2\sqrt2)bcd-bc-bd-cd+\lambda(b+c+d)=0$$ and similar or $$(4-2\sqrt2)abcd-abc-abd-acd+\lambda(ab+ac+ad)=0,$$ which gives $$-abc-abd-acd+\lambda(ab+ac+ad)=-abc-abd-bcd+\lambda(ab+bc+bd)$$ or $$(a-b)\left(\lambda(c+d)-cd\right)=0.$$ By the same way we can get that in the minimum point $(a,b,c,d,\lambda)$ holds: $$(a-c)\left(\lambda(b+d)-bd\right)=0,$$ $$(a-d)\left(\lambda(b+c)-bc\right)=0,$$ $$(b-c)\left(\lambda(a+d)-ad\right)=0,$$ $$(b-d)\left(\lambda(a+c)-ac\right)=0$$ and $$(c-d)\left(\lambda(a+b)-ab\right)=0.$$ 2.1 If $a=b=c=d$ then $a=1$ and the inequality is obviously true.

2.2 If $a\neq b$, $a\neq c$ and $a\neq d$ so we obtain: $$\lambda(c+d)-cd=0,$$ $$\lambda(b+d)-bd=0$$ and $$\lambda(b+c)-bc=0,$$ which gives $$\frac{bc}{b+c}=\frac{bd}{b+d}=\frac{cd}{c+d}$$ or $$\frac{1}{b}+\frac{1}{c}=\frac{1}{b}+\frac{1}{d}=\frac{1}{c}+\frac{1}{d}$$ or $$b=c=d.$$ Hence, the condition gives $a=\frac{2-b^2}{b}$, where $0<b<\sqrt2$ and we need to prove that $$(4-2\sqrt2)\cdot\frac{2-b^2}{b}\cdot b^3+2\sqrt2\geq \frac{2-b^2}{b}\cdot3b^2+b^3,$$ which is $$(b-1)^2(\sqrt2-b)(\sqrt2b+1+\sqrt2)\geq0;$$ 2.3 $a=b.$

If we want to get something new, we can get $a\neq c$ and $a\neq d$, which gives $$\frac{1}{b}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c}$$ or $$d=c.$$ Hence, the condition gives $a^2+4ac+c^2=6$ and we need to prove that $$(2-\sqrt2)a^2c^2+\sqrt2\geq(a+c)ac.$$ Let $a+c=2u$ and $ac=v^2$.

Hence, the condition gives $v^2+2u^2=3$ and since $u^2\geq v^2=3-2u^2$, we obtain $u\geq1$.

Id est, we need to prove that $$(2-\sqrt2)v^4+\sqrt2\geq2uv^2$$ or $$(2-\sqrt2)(3-2u^2)^2+\sqrt2\geq2u(3-2u^2)$$ or $$(u-1)((4-2\sqrt2)u^3+(6-2\sqrt2)u^2-(6-4\sqrt2)u-9+4\sqrt2)\geq0,$$ which is true for $u\geq1$.

Indeed, the polynomial $(4-2\sqrt2)u^3+(6-2\sqrt2)u^2-(6-4\sqrt2)u-9+4\sqrt2$

has one changing of coefficients signs, which by rule of Descartes says that

this polynomial has unique positive root.

Thus, $$ (4-2\sqrt2)u^3+(6-2\sqrt2)u^2-(6-4\sqrt2)u-9+4\sqrt2\geq$$ $$\geq(4-2\sqrt2)+(6-2\sqrt2)-(6-4\sqrt2)-9+4\sqrt2=4\sqrt2-5>0$$
Done!