Showing $\frac1{ab+4}+\frac1{ac+4}+\frac1{ad+4}+\frac1{bc+4}+\frac1{bd+4}+\frac1{cd+4}\geq\frac65$, for positive $a,b,c,d$ with $ab+bc+cd+da=4$

Question

Let us take $a\geq b\geq c\geq d>0$ such that $ab+bc+cd+da=4$. Show that $$\frac{1}{ab+4}+\frac{1}{ac+4}+\frac{1}{ad+4}+\frac{1}{bc+4}+\frac{1}{bd+4}+\frac{1}{cd+4}\geq\frac{6}{5}$$

• $a+b+c+d \ge 4$, $abcd \le 1$( are obvious)

I wasn't able to solve it using the following:

• using inequality of means.
• Using rearrangement inequality
• Substituting $\frac{1}{x}$ doesn't help since $f(x) = \frac{x}{1+4x}$ is concave.

Answer 1

Proof.

We have $ab, ac, ad, bc, bd, cd \le 4$. (Note: $ac + bd \le ab + bc \le 4$.)

Let $x = ab, y = bc, z = cd, u = da, v = ac, w = bd$. Then $x, y, z, u, v, w \le 4$, and $x + y + z + u = 4$.

We have $$\frac{1}{x + 4} = \frac25 - \frac{1}{25}(x + 4) + \frac{(x - 1)^2}{25(x + 4)} \ge \frac25 - \frac{1}{25}(x + 4) + \frac{(x - 1)^2}{25(4 + 4)}.$$ (Note: $\frac25 - \frac{1}{25}(x + 4)$ is the first order Taylor approximation of $\frac{1}{x+4}$ around $x = 1$.)
Similarly, we deal with $\frac{1}{y + 4}$ etc.

It suffices to prove that \begin{align*} &\frac{12}{5} - \frac{1}{25}(x + y + z + u + v + w + 24)\\[6pt] &\quad + \frac{(x - 1)^2 + (y-1)^2 + (z-1)^2 + (u-1)^2 + (v-1)^2 + (w-1)^2 }{25(4 + 4)}\\ \ge{}& \frac65, \end{align*} or \begin{align*} &\frac{27}{100} - \frac{1}{20}(x + y + z + u) + \frac{1}{200}(x^2 + z^2) + \frac{1}{200}(y^2 + u^2) \\[6pt] &\quad - \frac{1}{20}(v + w) + \frac{1}{200}(v^2 + w^2)\\[6pt] \ge{}& 0, \end{align*} or (using $xz = yu = vw = abcd$) \begin{align*} &\frac{27}{100} - \frac{1}{20}(x + y + z + u) + \frac{1}{200}(x + z)^2 + \frac{1}{200}(y + u)^2 \\[6pt] &\quad - \frac{1}{20}(v + w) + \frac{1}{200}(v + w)^2 - \frac{3}{100}abcd\\[6pt] \ge{}& 0. \tag{1} \end{align*}

Using $abcd \le \frac{1}{4}(ad + bc)^2 = \frac{1}{4}(y + u)^2$, and \begin{align*} &- \frac{1}{20}(v + w) + \frac{1}{200}(v + w)^2 - \left(- \frac{1}{20}(x + z) + \frac{1}{200}(x + z)^2\right)\\[6pt] ={}& \frac{1}{200}(x + z - v - w)(10 - x - z - v - w)\\[6pt] \ge{}& 0 \end{align*} (using $x + z - v - w = (a - c)(b - d) \ge 0$ and $x + z \le 4$), from (1), it suffices to prove that \begin{align*} &\frac{27}{100} - \frac{1}{20}(x + y + z + u) + \frac{1}{200}(x + z)^2 + \frac{1}{200}(y + u)^2 \\[6pt] &\quad - \frac{1}{20}(x + z) + \frac{1}{200}(x + z)^2 - \frac{3}{100}\cdot \frac{(y + u)^2}{4}\\[6pt] \ge{}& 0 \end{align*} or (using $(x + z) + (y + u) = 4$) $$\frac{3}{400}(x + z - 2)^2 \ge 0$$ which is clearly true.

We are done.