Gamma Function inequality .

Why does the following inequality hold :-

$$\sqrt 2 \frac {\Gamma ((n+1)/2)}{\Gamma (n/2)} -\sqrt 2 \frac {\Gamma ((m+1)/2)}{\Gamma (m/2)} \ge \sqrt n - \sqrt m$$

provided $n \ge m \ge 1$ .

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I suppose $n,m\in\mathbb{N}$ right? –  b00n heT Jan 28 at 16:10

Let $H(x)$ be the extended Harmonic Numbers $$H(x)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag{1}$$ and $$H'(x)=\sum_{k=1}^\infty\frac1{(k+x)^2}\tag{2}$$ Using this answer, we get \begin{align} &\frac{\mathrm{d}}{\mathrm{d}x}\log\left(\frac{\Gamma(x+1/2)}{\Gamma(x)}\right)\\ &=\sum_{k=1}^\infty\left(\frac1{k+x-1}-\frac1{k+x-1/2}\right)\\[6pt] &=H(x-1/2)-H(x-1)\tag{3} \end{align}

Theorem 1: $\dfrac{\Gamma(x+1/2)}{\Gamma(x)}$ is strictly concave for $x\gt-1/2$.

Proof: Since $H(x)$ is strictly concave, we have \begin{align} H(x-1/2)-H(x-1) &=H(x+1/2)-H(x)-\frac1{x+1/2}+\frac1x\\ &\lt\frac12(H(x+1/2)-H(x-1/2))+\frac1{x(2x+1)}\\ &=\frac{x+1}{x(2x+1)}\tag{4} \end{align} Furthermore, since $H'(x)$ is strictly convex, \begin{align} H'(x-1/2)-H'(x-1) &=H'(x+1/2)-H'(x)+\frac1{(x+1/2)^2}-\frac1{x^2}\\ &\lt\frac12(H'(x+1)-H'(x))-\frac{4x+1}{x^2(2x+1)^2}\\ &=-\frac1{2(x+1)^2}-\frac{4x+1}{x^2(2x+1)^2}\tag{5} \end{align} Using $(3)$ twice, then $(4)$ and $(5)$, says that for $x\gt-1/2$ \begin{align} &\frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\Gamma(x+1/2)}{\Gamma(x)}\\[6pt] &=\frac{\Gamma(x+1/2)}{\Gamma(x)}\left[\big(H(x-1/2)-H(x-1)\big)^2+H'(x-1/2)-H'(x-1)\right]\\ &\lt\frac{\Gamma(x+1/2)}{\Gamma(x)}\left[\frac{(x+1)^2}{x^2(2x+1)^2}-\frac1{2(x+1)^2}-\frac{4x+1}{x^2(2x+1)^2}\right]\\ &=-\frac{\Gamma(x+1/2)}{\Gamma(x)}\frac{2x^3+4x^2+7x+4}{2x(x+1)^2(2x+1)^2}\\[6pt] &\lt0\tag{6} \end{align} QED

Corollary 1: $\dfrac{\Gamma(x+1/2)}{\Gamma(x)}\gt\sqrt{x-1/4}$

Proof: $\dfrac{\Gamma(x+1/2)}{\Gamma(x)}$ is concave; therefore, \begin{align} \frac{\Gamma(x+1/2)^2}{\Gamma(x)^2} &\gt\frac12\left[\frac{\Gamma(x)}{\Gamma(x-1/2)}+\frac{\Gamma(x+1)}{\Gamma(x+1/2)}\right]\frac{\Gamma(x+1/2)}{\Gamma(x)}\\[6pt] &=x-1/4\tag{7} \end{align} QED

Theorem 2: $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\log\left(\frac{\Gamma(x+1/2)}{\Gamma(x)}\right)\gt\frac12\left(\frac1{x-1/4}-\frac1{12}\frac1{(x-1/4)^3}\right)$

Proof: Because $(k+x-3/4)^2=(k+x-1)(k+x-1/2)+\frac1{16}$, we have \begin{align} H(x-1/2)-H(x-1) &=\sum_{k=1}^\infty\left(\frac1{k+x-1}-\frac1{k+x-1/2}\right)\\ &=\frac12\sum_{k=1}^\infty\frac1{(k+x-1)(k+x-1/2)}\\ &\gt\frac12\sum_{k=1}^\infty\frac1{(k+x-3/4)^2}\\ &=\frac12H'(x-3/4)\tag{8} \end{align} As long as $x\gt1/2$, \begin{align} &\left(\frac1{x-1/2}-\frac1{x+1/2}\right)-\frac1{12}\left(\frac1{(x-1/2)^3}-\frac1{(x+1/2)^3}\right)\\ &=\frac1{x^2-1/4}-\frac1{12}\frac{3x^2+1/4}{(x^2-1/4)^3}\\ &=\frac{12(x^2-1/4)^2-3(x^2-1/4)-1}{12(x^2-1/4)^3}\frac{4(x^2-1/4)+1}{4x^2}\\ &=\frac{48(x^2-1/4)^3-7(x^2-1/4)-1}{48(x^2-1/4)^3}\frac1{x^2}\\ &\lt\frac1{x^2}\tag{9} \end{align} Using $(9)$ in $(2)$, we get that for $x\gt-1/2$, $$H'(x)\gt\frac1{x+1/2}-\frac1{12}\frac1{(x+1/2)^3}\tag{10}$$ Combining $(3)$, $(8)$, and $(10)$ proves the theorem.

QED

Corollary 2: $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\frac{\Gamma(x+1/2)}{\Gamma(x)}\gt\frac12\left(\frac1{\sqrt{x-1/4}}-\frac1{12}\frac1{\sqrt{x-1/4}^5}\right)$

Proof: multiply the results of Corollary 1 and Theorem 2.

QED

Corollary 3: $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\frac{\Gamma(x+1/2)}{\Gamma(x)}\gt\frac1{2\sqrt{x}}$

Proof: For $x\ge16/15$, $$\frac1{\sqrt{x-1/4}}-\frac1{12}\frac1{\sqrt{x-1/4}^5}\gt\frac1{\sqrt{x}}\tag{11}$$ Corollary 2 in conjunction with $(11)$ proves Corollary 3 for $x\ge16/15$.

QED

The proposition in Corollary 3 is valid for $x\ge1/5$, but the proof above only works for $x\ge16/15$.

The Mean Value Theorem says that there is a $z$ between $x$ and $y$ so that $$\frac{\frac{\Gamma(x+1/2)}{\Gamma(x)}-\frac{\Gamma(y+1/2)}{\Gamma(y)}}{\sqrt{x}-\sqrt{y}} =\frac{\frac{\mathrm{d}}{\mathrm{d}z}\frac{\Gamma(z+1/2)}{\Gamma(z)}}{\frac1{2\sqrt{z}}}\gt1\tag{12}$$ The inequality in $(12)$ is simply Corollary 3. Plug $x=n/2$ and $y=m/2$ into $(12)$ and you get the inequality in the question.