Inequality by Calculus

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Inequality by Calculus

Proof 1

If $x\in [a,b]\Rightarrow x^n\in [a^n,b^n]\;$ then

$\displaystyle\begin{align} &(x^n-a^n)(x^n-b^n)\leq 0\\ &x^{2n}-x^n(a^n+b^n)+a^nb^n\leq 0\\ &x^n-(a^n+b^n)+\frac{(ab)^n}{x^n}\leq 0\\ &x^n+\frac{(ab)^n}{x^n}\leq a^n+b^n \overbrace{\leq}^{\text{AM-PM}}\sqrt{2(a^{2n}+b^{2n})}\\ &x^n+\frac{(ab)^n}{x^n}\leq \sqrt{2(a^{2n}+b^{2n})}\\ &\int_a^b x^n dx+(ab)^n \int_a^b \frac{1}{x^n}dx\leq \sqrt{2(a^{2n}+b^{2n})}\int_a^b dx\\ &\frac{x^{n+1}}{n+1}\Bigg|_a^b+(ab)^n\frac{x^{-n+1}}{-n+1}\Bigg|_a^b\leq \sqrt{2(a^{2n}+b^{2n})}(b-a)\\ &\frac{1}{n+1}(b^{n+1}-a^{n+1})+(ab)^n\frac{1}{-n+1}\left(\frac{1}{b^{n-1}}-\frac{1}{a^{n-1}}\right)\leq \sqrt{2(a^{2n}+b^{2n})}(b-a)\\ &\frac{b^{n+1}-a^{n+1}}{n+1}+\frac{(ab)^n}{n-1}\left(\frac{a}{a^n}-\frac{b}{b^{n}}\right)\leq \sqrt{2(a^{2n}+b^{2n})}(b-a)\\ &\frac{b^{n+1}-a^{n+1}}{n+1}+\frac{(ab)^n}{n-1}\cdot \frac{ab^n-ba^n}{(ab)^n}\leq (b-a)\sqrt{2(a^{2n}+b^{2n})}\\ &\frac{b^{n+1}-a^{n+1}}{n+1}+\frac{ab(b^{n-1}-a^{n-1})}{n-1}\leq (b-a)\sqrt{2(a^{2n}+b^{2n})} \end{align}$

Proof 2

Since $0\lt a\leq b$, we consider separately two cases: $0\lt a=b\;$ and $0\lt a\lt b.$

When $a=b$ there is nothing to prove.

For the case where $a\lt b,\;$ we first prove the following lemma:

Lemma

$\forall x\in (1,\infty); n\in \mathbb{N}; n\geq 2, \displaystyle\frac{(1+x^n)(x-1)}{2x}>\frac{x^{n-1}-1}{n-1}.$

Proof of Lemma

Consider $\displaystyle f(x)=\frac{(1+x^n)(x-1)}{2x}-\int_a^x t^{n-2}dt,\;$ $x\geq 1,\;$ $n\in \mathbb{N},\;$ $n\geq 2.\;$ It is easy to show that $f(x)\;$ is continuous on $[1,\infty)\;$ and differentiable on $(1,\infty).\;$ Differentiating $f(x)\;$ gives

$f'(x)=\displaystyle\frac{nx^{n+1}+1-(n+1)x^n}{2x^2}.$

Now if $x>1\;$ then $x^n\gt x^i\;$ for $i=n-1, n-2, n-3, \ldots, 3,2,1,0$

$\begin{align}&\Rightarrow nx^n\gt x^{n-1}+x^{n-2}+x^{n-3}+\ldots +x^2+x+1=\frac{x^n-1}{x-1}\\ &\Rightarrow x^n-1 \lt nx^n(x-1)\\ &\Rightarrow 0\lt nx^{n+1}-nx^n+1-x^n=nx^{n+1}+1-(n+1)x^n\\ &\Rightarrow \frac{nx^{n+1}+1-(n+1)x^n}{2x^2}\gt 0. \end{align}$

It follows that

$f'(x)\gt 0 \text{ for } x\gt 1; n\in \mathbb{N}; n\geq 2.$

Thus $f(x)\;$ is increasing on $[1,\infty),\;$ implying $f(x)\gt f(1)$

$\displaystyle\begin{align} &\Rightarrow \frac{(1+x^n)(x-1)}{2x}-\int_1^x t^{n-2}dt\gt 0\\ &\Rightarrow \frac{(1+x^n)(x-1)}{2x}>\frac{x^{n-1}-1}{n-1},\;\forall x\in (1,\infty); n\in \mathbb{N}; n\geq 2. \end{align}$

And Lemma follows. Now

$\displaystyle\begin{align} \frac{b^{n+1}-a^{n+1}}{(b-a)(n+1)}&=\frac{b^n+ab^{n-1}+a^2b^{n-2}+\ldots+a^{n-1}b+a^n}{n+1}\\ &\lt \sqrt{\frac{b^{2n}+a^2b^{2n-2}+\ldots+a^{2n}}{n+1}}, \end{align}$

because of the AM-QM inequality. Similarly,

$\displaystyle\begin{align} &\frac{ab(b^{n-1}-a^{n-1})}{(b-a)(n-1)}=\frac{ab^{n-1}+a^2b^{n-2}+\ldots+a^{n-1}b}{n-1}\\ &\lt\sqrt{\frac{a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2}{n-1}}\\ &\Rightarrow \frac{b^{n+1}-a^{n+1}}{(b-a)(n+1)}+\frac{ab(b^{n-1}-a^{n-1})}{(b-a)(n-1)}\\ &\lt \sqrt{\frac{b^{2n}+a^2b^{2n-2}+\ldots +a^{2n}}{n+1}}+\sqrt{\frac{a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2}{n-1}} \end{align}$

Applying the AM-QM inequality again:

$\displaystyle\begin{align} &\sqrt{\frac{b^{2n}+a^2b^{2n-2}+\ldots+a^{2n}}{n+1}}+\sqrt{\frac{a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2}{n-1}}\\ &\lt\sqrt{2}\cdot \sqrt{\frac{b^{2n}+a^2b^{2n-2}+\ldots+a^{2n}}{n+1}+\frac{a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2}{n-1}} \end{align}$

Simplifying the right-hand side:

$\displaystyle\begin{align} &\frac{b^{2n}+a^2b^{2n-2}+\ldots+a^{2n}}{n+1}+\frac{a^2b^{2n-2}+a^4b^{2n-4}+\ldots a^{2n-2}b^2}{n-1}\\ &=\frac{(n-1)(a^{2n}+b^{2n})+2n(a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2)}{n^2-1}\\ &=\frac{a^{2n}+b^{2n}}{n+1}+\frac{2n}{n^2-1}\cdot a^2 b^{2n-2}\cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}} \end{align}$

(because $a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2=a^2b^{2n-2} \cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}}\;$ a Geometric Series)

So

$\displaystyle\begin{align} &\sqrt{\frac{b^{2n+a^2b^{n-2}+\ldots +a^{2n}}}{n-1}}+\sqrt{\frac{a^2b^{2n-2}+a^4b^{2n-4}+\ldots+a^{2n-2}b^2}{n-1}}\\ &\lt\sqrt{2}\cdot\sqrt{\frac{a^{2n}+b^{2n}}{n+1}+\frac{2n}{n^2-1}\cdot a^2b^{2n-2}\cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}}} \end{align}$

Now, since $0\lt a\lt b,\;$ we may use the above lemma with $x=\left(\displaystyle\frac{b}{a}\right)^2\gt 1$:

$\displaystyle\begin{align} &\frac{(1+x^n)(x-1)}{2x}\gt \frac{x^{n-1}-1}{n-1}\\ &\Rightarrow\frac{\left\{1+\left(\frac{b}{a}\right)^{2n}\right\}\left\{\left(\frac{b}{a}\right)^2-1\right\}}{2\left(\frac{b}{a}\right)^2}\gt\frac{(\frac{b}{a})^{2n-2}-1}{n-1}\\ &\Rightarrow\left\{1+\left(\frac{b}{a}\right)^{2n}\right\}\left\{1-\left(\frac{a}{b}\right)^2\right\}\gt\frac{2}{n-1}\cdot \left\{\left(\frac{b}{a}\right)^{2n-2}-1\right\}, \end{align}$

Multiplying both sides by $\displaystyle\frac{a^{2n}\cdot n}{n+1}$ gives:

$\displaystyle\begin{align} &\frac{n}{n+1}\cdot (a^{2n}+b^{2n})\cdot \left\{1-\left(\frac{a}{b}\right)^2\right\}\gt\frac{2n}{n^2-1}(a^2b^{2n-2}-a^{2n})\\ &\Rightarrow\left(1-\frac{1}{n+1}\right)\cdot (a^{2n}+b^{2n})\geq \frac{2n}{n^2-1}\cdot a^2b^{2n-2}\cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}}\\ &\Rightarrow a^{2n}+b^{2n}-\frac{a^{2n}+b^{2n}}{n+1}>\frac{2n}{n^2-1}\cdot a^2b^{2n-2}\cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}}\\ &\Rightarrow a^{2n}+b^{2n}>\frac{a^{2n}+b^{2n}}{n+1}+\frac{2n}{n^2-1}\cdot a^2 b^{2n-2}\cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}}\\ &\Rightarrow \sqrt{2}\sqrt{\frac{a^{2n}+b^{2n}}{n+1}+\frac{2n}{n^2-1}\cdot a^2b^{2n-2}\cdot \frac{1-(\frac{a^2}{b^2})^{n-1}}{1-\frac{a^2}{b^2}}}<\sqrt{2(a^{2n}+b^{2n})} \end{align}$

Combining everything, we finally get:

$\displaystyle\frac{b^{n+1}-a^{n+1}}{(b-a)(n+1)}+\frac{ab(b^{n-1}-a^{n-1})}{(b-a)(n-1)}\lt\sqrt{2(a^{2n}+b^{2n})}$

Multiplying both sides by $b-a>0$,

$\displaystyle \frac{b^{n+1}-a^{n+1}}{n+1}+\frac{ab(b^{n-1}-a^{n-1})}{n-1}\lt (b-a)\sqrt{2(a^{2n}+b^{2n})}$

for $0\lt a\lt b;\;$ $n\in \mathbb{N}; n\geq 2\;$ which furnishes the proof.

Proof 3

$LHS=\int_a^b (x^n+abx^{n-2})dx.\;$ By the AM-QM inequality, $(b-a)(b^n+a^n)\leq RHS.\;$ But

$\displaystyle (b-a)(b^n+a^n)=\int_a^b\left[(n+1)x^n-ab(n-1)x^{n-2}\right]dx.$

Hence suffice it to show that

$\displaystyle\begin{align}&\int_a^b (x^n+abx^{n-2})dx\leq \int_a^b \left[(n+1)x^n-ab(n-1)x^{n-2}\right]dx\\ &\Leftrightarrow\int_a^b\left[x^{n-2}(x^2-ab)\right]dx\geq 0. \end{align}$

But from Chebyshev's inequality,

$\displaystyle\begin{align} (b-a)\int_a^b \left[x^{n-2}(x^2-ab)\right]dx &\geq \left(\int_a^b x^{n-2}dx\right)\left(\int_a^b (x^2-ab)dx\right)\\ &=\left(\int_a^b x^{n-2}dx\right)\cdot \frac{(b-a)^3}{3}\geq 0 \end{align}$

So that

$\displaystyle\int_a^b \left[x^{n-2}(x^2-ab)\right]dx\geq 0.$

Proof 4

Use Hermite - Hadamard inequality:

For a convex and continuous function $f$

$\displaystyle\frac{\int_a^b f(x)dx}{b-a}\leq \frac{f(a)+f(b)}{2}.$

So let $f(x)=x^n.\;$ Then

$\displaystyle\frac{\int_a^b x^n dx}{b-a}\leq \frac{a^n+b^n}{2},$

or,

$\displaystyle\frac{b^{n+1}-a^{n+1}}{(n+1)(b-a)}\leq \frac{a^n+b^n}{2}.$

Similarly:

$\displaystyle\frac{b^{n-1}-a^{n-1}}{(n-1)(b-a)}\leq \frac{a^{n-2}+b^{n-2}}{2},$

implying

$\displaystyle\begin{align} ab\cdot \frac{b^{n-1}-a^{n-1}}{(n-1)(b-a)}&\leq ab\cdot \frac{a^{n-2}+b^{n-2}}{2}\\ &\leq \frac{a^2+b^2}{2}\cdot \frac{a^{n-2}+b^{n-2}}{2}\\ &\leq \frac{a^n+b^n}{2}. \end{align}$

Now use AM-GM and Chebyshev's inequality to obtain

$\displaystyle\begin{align}\frac{b^{n+1}-a^{n+1}}{(n+1)(b-a)}+ab\cdot \frac{b^{n-1}-a^{n-1}}{(n-1)(b-a)}&\leq \frac{a^n+b^n}{2}+\frac{a^n+b^n}{2}\\ &=a^n+b^n \end{align}$

However, with $a^n=x, b^n=y,\;$ $a^n+b^n \leq \sqrt{2(a^{2n}+b^{2n})}\;$ becomes $x+y\leq \sqrt{2(x^2+y^2)},\;$ i.e., $(x+y)^2\leq 2(x^2+y^2),\;$ or: $2xy\leq x^2+y^2,\;$ which is of course correct.

Acknowledgment

I owe my deepest gratitude to Daniel Sitaru who communicate to me the problem, along with four proofs, and kindly added the LaTeX files for all four. As SP.022, the problem has been published in the Romanian Mathematical Magazine, n 2, Autumn 2016.

The inequality and Proof 1 are due to Dan Sitaru (Romania); Proof 2 is by Hamza Mahmood (Pakistan); Proof 3 is by Leonard Giugiuc (Romania); Proof 4 is by Marian Dinca (Romania).

 

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