Prove that $(a+b)(b+c)(c+d)(d+a)\geq(a+1)(b+1)(c+1)(d+1)$

Question

Question:

Let $a, b, c, d$ be positive real numbers satisfying $abcd=1$. Prove that $(a+b)(b+c)(c+d)(d+a)\geq(a+1)(b+1)(c+1)(d+1)$. Characterise the instances of equality.

My (failed) approach: I immediately noticed that equality holds when $a=b=c=d=1$ and thought of using AM-GM Inequality. I tried $(a+b)(b+c)(c+d)(d+a)\geq16abcd=16$ (By applying AM-GM Inequality on each term). Hence, it suffices to prove $16\geq(a+1)(b+1)(c+1)(d+1)$, but this is not true, and I got stuck. Any help will be appreciated.

Answer 1

The plan is to show:

$$\frac{(a + 1)(b + 1)(c + 1)(d + 1)}{(a+b)(b+c)(c + d)(d+a)} \leq 1$$

We first split up the expression in a way that is more suitable for AM-GM:

$$\frac{(a + 1)(b + 1)(c + 1)(d + 1)}{(a+b)(b+c)(c + d)(d+a)} = \frac{(a+1)(b+1)}{(a+b)} \frac{(c+1)(d+1)}{(c+d)} \frac{1}{(b+c)(d+a)}$$

Now taking a look at the single $\frac{(a+1)(b+1)}{(a+b)}$ part, we can apply AM-GM ($a + b \geq 2\sqrt{ab}$):

$$\begin{align*} \frac{(a+1)(b+1)}{(a+b)} &= \frac{ab + a + b + 1}{a + b} \\ &= \frac{ab}{a+b} + 1 + \frac{1}{a + b}\\ &\leq \frac{ab}{2\sqrt{ab}} + 1 + \frac{1}{2\sqrt{ab}} \\ &= \frac{ab + 2\sqrt{ab} + 1}{2\sqrt{ab}} \\ &= \frac{\left(\sqrt{ab} + 1\right)^2}{2\sqrt{ab}} \\ \end{align*}$$

Now, we plug this into our main equation:

$$\begin{align*} \frac{(a+1)(b+1)}{(a+b)} \frac{(c+1)(d+1)}{(c+d)} \frac{1}{(b+c)(d+a)} &\leq \frac{\left(\sqrt{ab} + 1\right)^2}{2\sqrt{ab}} \frac{\left(\sqrt{cd} + 1\right)^2}{2\sqrt{cd}} \frac{1}{(b+c)(d+a)} \\ &= \frac{\left(\sqrt{ab} + 1\right)^2 \left(\sqrt{cd} + 1\right)^2}{4\sqrt{ab}\sqrt{cd}(b+c)(d+a)} \\ &= \frac{\left(\sqrt{ab} + 1\right)^2 \left(\sqrt{cd} + 1\right)^2}{4(b+c)(d+a)} \end{align*}$$

Then using Cauchy-Schwarz: $$|\vec{p}||\vec{q}| \geq |\vec{p} \cdot \vec{q}|^2 \implies (p_x^2 + p_y^2)(q_x^2 + q_y^2) \geq (p_x q_x + p_y q_y)^2$$

On the denominator $(b + c)(d + a)$ to get:

$$(b + c)(d + a) = (\sqrt{b}^2 + \sqrt{c}^2)(\sqrt{a}^2 + \sqrt{d}^2) \geq (\sqrt{ab} + \sqrt{cd})^2$$

Therefore we can decrease the denominator and get:

$$\begin{align*} \frac{\left(\sqrt{ab} + 1\right)^2 \left(\sqrt{cd} + 1\right)^2}{4(b+c)(d+a)} &\leq \frac{\left(\sqrt{ab} + 1\right)^2\left(\sqrt{cd} + 1\right)^2}{4(\sqrt{ab} + \sqrt{cd})^2} \\ &= \frac{1}{4} \left( \frac{\left(\sqrt{ab} + 1\right) \left(\sqrt{cd} + 1\right)}{(\sqrt{ab} + \sqrt{cd})} \right)^2 \\ &= \frac{1}{4} \left( \frac{ \sqrt{abcd} + \sqrt{ab} + \sqrt{cd} + 1 }{\sqrt{ab} + \sqrt{cd}} \right)^2 \\ &= \frac{1}{4} \left( \frac{ \sqrt{1} + 1 }{\sqrt{ab} + \sqrt{cd}} + 1\right)^2 \\ &= \frac{1}{4} \left( \frac{ 2 }{\sqrt{ab} + \sqrt{cd}} + 1\right)^2 \\ &= \frac{1}{4} \left( \frac{ 2 }{2\sqrt[4]{abcd}} + 1\right)^2 \text{ by AM-GM again} \\ &= \frac{1}{4} \left( 1 + 1\right)^2 \leq 1 \end{align*}$$

Answer 2

Let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2,$ where $v>0$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4,$ where $t>0$.

Thus, since $$\prod_{cyc}(a+b)-\sum_{cyc}a\sum_{cyc}abc=(ac-bd)^2\geq0,$$ it's enough to prove that: $$16uw^3\geq\prod_{cyc}(a+1)$$ or $$8uw^3\geq t^4+2ut^3+3v^2t^2+2w^3t,$$ which is true because by Maclaurin $u\geq v\geq w\geq t$ and by Rolle $w^6\geq v^2t^4.$