Dan Sitaru's Inequality by Induction


Dan Sitaru's Inequality by Induction

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

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


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

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.


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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