$a+b+c+d=1, a,b,c,d ≠ 0$, then prove that $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 + (c + \frac{1}{c})^2 + (d + \frac{1}{d})^2 \ge \frac{289}{4} $

Question

If $a+b+c+d=1$, $a,b,c,d ≠ 0$, prove that $$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 + \left(c + \frac{1}{c}\right)^2 + \left(d + \frac{1}{d}\right)^2 \ge \frac{289}{4} $$

I tried expanding the entire LHS and I got $a^2 + b^2 + c^2 + d^2 + \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + 8$ on the LHS. Then I tried applying AM-GM to the first 4 terms and the last 4 terms and I got $4 \sqrt{abcd} + \frac{4}{\sqrt{abcd}}$ but that didnt seem useful.

Any hints? Thank you in advance.

Answer 1

Let $$f(x)=x^2+\frac{1}{x^2}.$$

Thus, $f$ is a convex function on $(0,+\infty)$ and $f$ is a convex function on $(-\infty,0)$.

We need to consider the following cases.

1. $a\geq b\geq c\geq d>0$.

The case $0>a\geq b\geq c\geq d$ is impossible because $a+b+c+d=1.$

Now, our inequality follows from the Jensen's inequality: $$8+\sum_{\text{cyc}}\left(a^2+\frac{1}{a^2}\right)\geq8+4\left(\left(\frac{a+b+c+d}{4}\right)^2+\frac{1}{\left(\frac{a+b+c+d}{4}\right)^2}\right)=\frac{289}{4}.$$

2. $a\geq b\geq c>0>d$ (the case $a>0>b\geq c\geq d$ is a similar).

3. $a\geq b>0>c\geq d$.

In the last case we obtain a counterexample: $$(a,b,c,d)=\left(-1,-1,\frac{3}{2},\frac{3}{2}\right).$$ In this case $$\sum_{\text{cyc}}\left(a+\frac{1}{a}\right)^2=\frac{313}{18}<\frac{289}{4},$$ which says that your inequality is wrong!

Answer 2

If you consider $a,b,c,d$ are all positive According to cauchy-schwarz inequality $(1+1+1+1)*((a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2)>=$ $(a+b+c+d+1/a+1/b+1/c+1/d)^2=$ $(a+b+c+d+4+b/a+a/b+c/a+a/c+d/a+a/d+c/d+d/c+c/b+b/c+d/b+b/d)^2>=(5+12)^2$[because $a/b+b/a>=2 and a+b+c+d=1$] So we get.... $(a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2>= 289/4$.