# Dan Sitaru's Inequality by Induction

### Problem

### Solution 1

Let $P(n)\,$ be the statement:

If $x_k\ge 1,\,$ $k=1,\ldots,n,\,$ then

$x_1+x_2+\ldots+x_n\leq n-1+x_1x_2\ldots x_n.$

We prove $P(n)\,$ by induction.

For $n=1,\,$ $x_1\ge 1-1+x_1,\,$ which is obviously true.

For $n=2,\,$ $x_1+x_2\le 2-1+x_1x_2\,$ is equivalent to $(x_1-1)(x_2-1)\ge 0,\,$ which is true because $x_1,x_2\ge 1.$

Now assume that for $n=k\ge 2,\,$ $P(k)\,$ is true and let there be numbers $x_1,\ldots,x_k,x_{k+1}\ge 1.\,$ We wish to prove that $P(k+1)\,$ is true, i.e.,

$x_1+x_2+\ldots+x_k+x_{k+1}\leq k+x_1x_2\ldots x_k x_{k+1}$

holds. By the inductive assumtion.

$x_1+x_2+\ldots+x_k\leq k-1+x_1x_2\ldots x_k$

and, therefore,

$x_1+x_2+\ldots+x_k+x_{k+1}\leq k-1+x_1x_2\ldots x_k+x_{k+1}.$

Suffice it to prove that

$k-1+x_1x_2\ldots x_k+x_{k+1}\leq k+x_1x_2\ldots x_k x_{k+1},$

i.e.,

$x_1x_2\ldots x_k+x_{k+1}\leq 1+x_1x_2\ldots x_k x_{k+1}.$

But this is equivalent to $0\leq x_1x_2\ldots x_k(x_{k+1}-1)-(x_{k+1}-1),\,$ i.e.,

$(x_1x_2\ldots x_k-1)(x_{k+1}-1)\geq 0,$

which is obviuosly true. This completes the induction.

Now, in particular, for $n=2,3,4,\,$ we have

$\begin{align} x_1+x_2&\leq 1+x_1x_2\\ x_1+x_2+x_3&\leq 2+x_1 x_2 x_3\\ x_1+x_2+x_3+x_4&\leq 3+x_1x_2x_3x_4\\ \end{align}$

By adding:

$3x_1+3x_2+2x_3+x_4\leq 6+x_1x_2+x_1x_2x_3+x_1x_2x_3x_4$

Setting $x_1=t^a; x_2=t^b; x_3=t^c; x_4=t^d$ leads to

$3t^a+3t^b+2t^c+t^d\leq 6+t^{a+b}+t^{a+b+c}+t^{a+b+c+d}.$

The inequality is preserved after integration:

$\displaystyle\begin{align} &3\int_0^1 t^a dt+3\int_0^1 t^b dt+2\int_0^1 t^c dt +\int_0^1 t^d dt\\ &\qquad\qquad\leq 6+\int_0^1 t^{a+b}dt +\int_0^1 t^{a+b+c}dt+\int_0^1 t^{a+b+c+d}dt \end{align}$

which is the required inequality

$\displaystyle\begin{align}&\frac{3}{a+1}+\frac{3}{b+1}+\frac{2}{c+1}+\frac{1}{d+1}\\ &\qquad\le 6+\frac{1}{a+b+1}+\frac{1}{a+b+c+1}+\frac{1}{a+b+c+d+1}. \end{align}$

### Solution 2

$\displaystyle \begin{align} &&I_1:\int_0^b \left[\frac{1}{(x+1)^2}-\frac{1}{(a+x+1)^2}\right]dx \geq 0 \\ &&I_2:\int_0^c \left[\frac{1}{(x+1)^2}-\frac{1}{(a+b+x+1)^2}\right]dx \geq 0 \\ &&I_3:\int_0^d \left[\frac{1}{(x+1)^2}-\frac{1}{(a+b+c+x+1)^2}\right]dx \geq 0 \end{align}$

The above inequalities follow from the integrands and the upper integration limits being non-negative. Carrying out the integration,

$\displaystyle \begin{align} &&I_1:-\frac{1}{b+1}+1+\frac{1}{a+b+1}-\frac{1}{a+1} \geq 0 \\ &&I_2:-\frac{1}{c+1}+1+\frac{1}{a+b+c+1}-\frac{1}{a+b+1} \geq 0 \\ &&I_3:-\frac{1}{d+1}+1+\frac{1}{a+b+c+d+1}-\frac{1}{a+b+c+1} \geq 0 \end{align}$

$3I_1+2I_2+I_3$ simplifies to the inequality to be proven.

### Acknowledgment

This problem from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. He later mailed his solution (Solution 1) in a LaTeX file as did Amit Itagi (Solution 2).

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