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The problem is:

Set $x_1,x_2, ..., x_n >0$, and $x_1x_2\cdots x_n=1$. Prove that $$\frac{1}{x_1(1+x_1)}+ \frac{1}{x_2(1+x_2)}+\cdots+\frac{1}{x_n(1+x_n)}\geqslant \frac{n}{2}.$$

I have already look at the solution but still very confused, the first line said that

OBVIOUSLY, the original inequality is equal to $$\frac{1+x_1+x_1^2}{x_1(1+x_1)}+\frac{1+x_2+x_2^2}{x_2(1+x_2)}+\cdots+\frac{1+x_n+x_n^2}{x_n(1+x_n)}\geqslant3n/2.$$

I am sorry that I am a dumb person, but I really can not find it is in any aspects "obvious", so i just calculate it:
since it said that "it is equal to", so I assume that it means that
1/3($\frac{1}{x_1(1+x_1)}+ \frac{1}{x_2(1+x_2)}+...+\frac{1}{x_n(1+x_n)}$)= $\frac{1+x_1+x_1^2}{x_1(1+x_1)}+\frac{1+x_2+x_2^2}{x2(1+x_2)}+...+\frac{1+x_n+x_n^2}{x_n(1+x_n)}$
by simplifing the above equation I got that:
$1+x_i+x_i^2=3$, and I am stucked at here can not make any steps further.

user26857
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Tong Su
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1 Answers1

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Observe that we simply added $1$ to each of the $n$ fractions:

$$\frac{1}{x_k(1+x_k)} + 1 = \frac{1+x_k(1+x_k)}{x_k(1+x_k)} = \frac{1 + x_k + x_k^2}{x_k(1+x_k)}$$

There are $n$ of these fractions. Hence, doing this to the left hand side of the inequality is tantamount to adding $n$ to the right hand side of the inequality.

PrincessEev
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