Inequalities of Jensen and Karamata

We will prove this using the method of mathematical induction. For \(n=2\) the statement follows from the convexity of the function. Assume that the statement holds for some \(n\geq 2\). Assume that we have two sequences of real numbers \(x_1, \dots, x_{n+1}\) and \(\alpha_1, \dots, \alpha_{n+1}\). Let \(A=\alpha_1+\cdots+ \alpha_n\). Then \begin{eqnarray*}f(\alpha_1x_1+\cdots +\alpha_nx_n+\alpha_{n+1}x_{n+1})&=& f\left(A\cdot\left(\frac{\alpha_1}Ax_1+\cdots+\frac{\alpha_n}Ax_n\right)+\alpha_{n+1}x_{n+1}\right)\\&\leq & Af\left(\frac{\alpha_1}Ax_1+\cdots+\frac{\alpha_n}Ax_n\right)+\alpha_{n+1}x_{n+1}\\ &\leq & A\left(\frac{\alpha_1}Af(x_1)+\cdots + \frac{\alpha_n}Af(x_n)\right)+\alpha_{n+1}x_{n+1}\\ &=&\alpha_1f(x_1)+\cdots+ \alpha_nf(x_n)+\cdots+\alpha_{n+1}f(x_{n+1}). \end{eqnarray*} The equality holds if and only if \(x_1=\cdots=x_n\).

Taking the logarithm of both left and right side the inequality becomes equivalent to \[\frac{x_1+\cdots+x_n}n\log\left(\frac{x_1+\cdots+x_n}n\right)\leq \frac1nx_1\log x_1+\cdots + \frac1nx_n\log x_n .\] This follows from the Jensen’s inequality applied to the function \(f(x)=x\log x\) and the sequence \(\alpha_1=\cdots=\alpha_n=\frac1n\).

From \(\frac1{1+xy}=\frac z{z+xyz}=\frac{z}{x+y+z+z}=\frac{z}{S+z}\), where \(S=x+y+z\). The function \(f(t)=\frac{t}{S+t}=1-\frac1{S+t}\) is concave hence \(f\left(\frac{x+y+z}3\right)\geq \frac{f(x)+f(y)+f(z)}3\). Since \(f\left(\frac{x+y+z}3\right)= f\left(\frac S3 \right)=\frac{\frac S3}{S+\frac S3}=\frac14\) the required statement follows from Jensen’s inequality.

The left-hand side can be written as \(f(x)+f(y)+f(z)\) for \(f(t)=\frac{t^{3/2}}{S-t}\), where \(S=x+y+z\). For every \(S\), the function \(f\) is concave on \((0,S)\). This function is convex for every \(S\). Indeed, \(f^{\prime}(t)=\frac{\frac32 t^{1/2}(S-t)+t^{3/2}}{(S-t)^2}=\frac{t^{1/2}}{2(S-t)^2}\cdot (3S-t)\). Then \begin{eqnarray*}f’’(t)&=&\frac{\left(\frac12t^{-1/2}(3S-t)-t^{1/2}\right)(S-t)+t^{1/2}(3S-t)\cdot2 }{2(S-t)^3}\\ &=&\frac{\left(3S-t-2t\right)(S-t)+4(3S-t)}{4t^{1/2}(S-t)^3}\geq 0, \end{eqnarray*} for \(t\in(0,S)\). Therefore \[f(x)+f(y)+f(z)\geq 3f\left(\frac{x+y+z}3\right)=3f(S/3)=3 \frac{\frac S3\sqrt{\frac S3}}{S-\frac S3}= 3\cdot \sqrt {\frac S3}\cdot \frac12\geq 3\cdot\frac1{2\sqrt 3}=\frac{\sqrt 3}2.\]

Let \(c_i=\frac{f(y_i)-f(x_i)}{y_i-x_i}\), for \(y_i\neq x_i\), and \(c_i=f^{\prime}_{+}(x_i)\), for \(x_i=y_i\). Since \(f\) is convex, and \(x_i\), \(y_i\) are decreasing sequences, \(c_i\) is non-increasing (because is represents the slope of \(f\) on the interval between \(x_i\) and \(y_i\)). We now have \begin{eqnarray*} \sum_{i=1}^n f(x_i)- \sum_{i=1}^n f(y_i)&=&\sum_{i=1}^n c_i(x_i-y_i)=\sum_{i=1}^nc_ix_i-\sum_{i=1}^n c_iy_i\\ &=&\sum_{i=1}^n(c_i-c_{i+1})(x_1+\cdots +x_i) \\&&- \sum_{i=1}^n(c_i-c_{i+1})(y_1+\cdots +y_i), \quad\quad\quad\quad\quad (2) \end{eqnarray*} here we define \(c_{n+1}\) to be \(0\). Now, denoting \(A_i=x_1+\cdots + x_i\) and \(B_i=y_1+\cdots + y_i\), (2) can be rearranged to \begin{eqnarray*} \sum_{i=1}^n f(x_i)- \sum_{i=1}^n f(y_i) &=& \sum_{i=1}^{n-1} (c_i-c_{i+1}) (A_i-B_i) + c_n\cdot (A_n-B_n). \end{eqnarray*} The sum on the right-hand side of the last inequality is non-negative because \(c_i\) is decreasing and \(A_i\geq B_i\). The last term \(c_n(A_n-B_n)\) is zero under the assumption (a). Under the assumption (b) we have that \(c_n\geq 0\) (\(f\) is increasing) and \(A_n\geq B_n\) and this implies (1).

Let \(a_i=e^{x_i}\) and \(b_i=e^{y_i}\). We easily verify that the conditions (b) of the Karamata’s theorem are satisfied. Thus \(\sum_{i=1}^n e^{y_i} \geq \sum_{i=1}^ne^{x_i}\) and the result immediately follows.

Rearrange \((2x_1-x_2, 2x_2-x_3,\dots, 2x_n-x_1)\) and \((x_1, \dots, x_n)\) in two non-increasing sequences \((2x_{m_1}-x_{m_1+1}, 2x_{m_2}-x_{m_2+1}, \dots, 2x_{m_n}-x_{m_n+1})\) and \((x_{k_1}, x_{k_2}, \dots, x_{k_n})\) (here we assume that \(x_{n+1}=x_1\). We will verify that condition (a) of the Karamata’s inequality is satisfied, and that follows from \begin{eqnarray}\nonumber &&(2x_{m_1}-x_{m_1+1}+ \cdots+ 2x_{m_l}-x_{m_l+1})-(x_{k_1}+\cdots + x_{k_l}) \\ \nonumber &\geq & (2x_{k_1}-x_{k_1+1}+ \cdots+ 2x_{k_l}-x_{k_l+1})-(x_{k_1}+\cdots + x_{k_l}) \\ &=& (x_{k_1}+\cdots x_{k_l})-(x_{k_1+1}+\cdots + x_{k_l+1})\geq 0.\nonumber \end{eqnarray} The function \(f(x)=-\cos x\) is convex on \([-\pi/2,\pi/2]\) hence Karamata’s inequality holds and we get \[-\cos(2x_1-x_2)-\cdots -\cos(2x_n-x_1)\geq -\cos x_1-\cdots - \cos x_n,\] which is obviously equivalent to the required inequality.

Choose \(x_i\) such that \(a_i=e^{x_i}\). When the sequences \((2x_1-x_2, \dots, 2x_n-x_1)\) and \((x_1, \dots, x_n)\) are sorten in non-increasing order, the first majorizes the second, and Karamata’s inequality applied to the convex function \(f(x)=1+e^x\) yields the desired result.