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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.

amWhy
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Aaron
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2 Answers2

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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: $$8+\sum_{\text{cyc}}\left(a^2+\frac{1}{a^2}\right)\geq8+4\left(\left(\frac{a+b+c+d}{4}\right)^2+\frac{1}{\left(\frac{a+b+c+d}{4}\right)^2}\right)=\frac{289}{4}.$$

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

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

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!

sai-kartik
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Michael Rozenberg
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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.... $(a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2>= 289/4$.

Michael Rozenberg
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Sayan
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  • If any one has any doubt then ask I will answer – Sayan May 14 '20 at 05:50
  • $\frac{a}{b}+\frac{b}{a}\geq2$ is wrong. Try $a=-1$ and $b=1$. – Michael Rozenberg May 14 '20 at 06:21
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    I think the question is incomplete if he will say a,b,c,d are all positive then my method will be right....it is common sense that my method is AM-GM inequality..also my fault not to read the problem properly.... thanks – Sayan May 14 '20 at 06:36
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    Could you please take the time to typeset your equation using mathjax? This will make your answer more easy to read and will rule out any ambiguities that may arise. – sai-kartik May 14 '20 at 06:38
  • Now I see your solution it is good... – Sayan May 14 '20 at 06:39
  • sai-kartik actually I am new here so I only use my mobile keyboard I don't know how to do easily.. thank you – Sayan May 14 '20 at 06:43
  • Michael Rozenberg can you please edit it??? – Sayan May 14 '20 at 06:46
  • @Sayan You can-not assume that $a$, $b$, $c$ and $d$ are positives if it's not given. You just solved another problem. – Michael Rozenberg May 14 '20 at 07:39
  • Hmm Michael Rozenberg from your solution the question is wrong when a, b, c, d are allowed to be negative, so I guess the question can be modified to only allow a, b, c, d be positive? I got this question from a friend, I don't know if it is correct. – Aaron May 14 '20 at 07:47
  • Michael Rozenberg thank you so much – Sayan May 14 '20 at 15:41