Problem 1. Given two poistive numbers $x, y.$ Prove that $$\left | x\ln x- y\ln y \right |\leq\sqrt[3]{\left | x^{2}- y^{2} \right |}+ \sqrt[3]{\left | x^{2}- y^{2} \right |^{2}}$$
There exists a solution using the integral to solve, without loss of generality, assume that $x\geq y\Rightarrow t= x^{2}- y^{2}\geq 0,$ let $b= y^{2},$ we need to prove the following integral inequality in 2 cases : $b\geq\frac{1}{e^{2}}$ either $b+ t\leq\frac{1}{e^{2}}$ $$\left | \int_{b}^{b+ t}\frac{\ln x+ 2}{4\sqrt{x}}{\rm d}x \right |\leq\sqrt[3]{t}+ \sqrt[3]{t^{2}}$$ Case 1. $b\geq\frac{1}{e^{2}}$ $$\int_{b}^{b+ t}\frac{\ln x+ 2}{4\sqrt{x}}{\rm d}x\leq\sqrt[3]{t}+ \sqrt[3]{t^{2}}\Leftrightarrow\int_{0}^{t}\frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\leq\int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}{\rm d}x$$ Substitute $g\left ( b \right )= \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}\Rightarrow {g}'\left ( b \right )= -\frac{\ln\left ( x+ b \right )}{8\left ( x+ b \right )^{\frac{3}{2}}}$
If $t\geq 1- \frac{1}{e^{2}}$ $$H= \int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x= \int_{0}^{1- \frac{1}{e^{2}}}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x+$$ $$+ \int_{1- \frac{1}{e^{2}}}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\geq\int_{0}^{1- \frac{1}{e^{2}}}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x+$$ $$\int_{1- \frac{1}{e^{2}}}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}}{\rm d}x= \int_{1- \frac{1}{e^{2}}}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}}{\rm d}x$$ $$+ \frac{1}{2}\sqrt{b}\ln b+ \left ( 1- e^{-2} \right )^{-\frac{1}{3}}+ \left ( 1- e^{-2} \right )^{-\frac{2}{3}}- \frac{1}{2}\sqrt{b+ 1- e^{-2}}\ln\left ( b+ 1- e^{-2} \right )$$ Let $f\left ( x \right )= \frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}},$ since $f\left ( x \right )\leq 0\,{\rm as}\,36.29678351< x< 1161.52586,$ $$f\left ( x \right )\geq 0\,{\rm as}\,1- e^{-2}< x< 36.29678351, 1161.52586\leq x$$ $$\Rightarrow H= \frac{1}{2}\sqrt{b}\ln b+ \left ( 1- e^{-2} \right )^{-\frac{1}{3}}+ \left ( 1- e^{-2} \right )^{-\frac{2}{3}}- \frac{1}{2}\sqrt{b+ 1- e^{-2}}\ln\left ( b+ 1- e^{-2} \right )+$$ $$+ \int_{1- \frac{1}{e^{2}}}^{1161.52586}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ \frac{1}{e^{2}} \right )+ 2}{4\sqrt{x+ \frac{1}{e^{2}}}}{\rm d}x\geq 1.431081083- 1.123709381> 0$$ If $t\leq 1- e^{-2}$ $$H= \int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\geq\int_{0}^{t}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}- \frac{1}{2}{\rm d}x\geq 0$$ Case 2. $b+ t\leq\frac{1}{e^{2}}$ $$H= \int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}+ \frac{\ln\left ( x+ b \right )+ 2}{4\sqrt{x+ b}}{\rm d}x\geq\int_{0}^{t}\frac{1}{3}x^{-\frac{2}{3}}+ \frac{2}{3}x^{-\frac{1}{3}}+ \frac{\ln x+ 2}{4\sqrt{x}}{\rm d}x\geq$$ $$\geq -\frac{1}{2}\sqrt{t}\ln t+ t^{\frac{1}{3}}+ t^{\frac{2}{3}}\geq 0$$
Problem 2. Given two real numbers $x, y.$ Prove that $$\left | \sin x- \sin y \right |\leq\left | x- y \right |$$
I'm having trouble using the way that we used in problem 1, I don't know how to devide it into some cases. On the other hand does there exist a nicer solution for problem 1 ; and another problem(s) as same as these ones as above, thank you.
Alternative solution:
First we give some auxiliary results (Facts 1-2). The proofs are given at the end.
Fact 1: Let $t > 1$ and $u > 0$. Then $$(t^2-1)^{1/3}u^{-1} + (t^2 - 1)^{2/3}u - t\ln t - 3(t-1)\ln u \ge 0.$$
Fact 2: Let $t > 1$ and $u > 0$. Then $$(t^2-1)^{1/3}u^{-1} + (t^2 - 1)^{2/3}u + t\ln t + 3(t-1)\ln u \ge 0.$$
Now, WLOG, assume $x > y$.
Let $t = \frac{x}{y}$. Then $t > 1$. The desired inequality is written as $$|t\ln t + (t-1)\ln y | \le (t^2-1)^{1/3}y^{-1/3} + (t^2 - 1)^{2/3}y^{1/3}. \tag{1}$$ It suffices to prove that $$t\ln t + (t-1)\ln y \le (t^2-1)^{1/3}y^{-1/3} + (t^2 - 1)^{2/3}y^{1/3}, \tag{2}$$ and $$- t\ln t - (t-1)\ln y \le (t^2-1)^{1/3}y^{-1/3} + (t^2 - 1)^{2/3}y^{1/3}. \tag{3}$$ By Facts 1-2, (2) and (3) are both true.
We are done.
$\phantom{2}$
Proof of Fact 1: Denote LHS by $f(u)$. We have \begin{align} f'(u) &= - (t^2-1)^{1/3}u^{-2} + (t^2 - 1)^{2/3} - 3(t-1)u^{-1}\\ &= \frac{1}{u^2}(u - u_0)\Big[(t^2-1)^{2/3}u + (t^2-1)^{1/3} u_0^{-1}\Big] \end{align} where $u_0 = \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2(t^2 - 1)^{2/3}}$.
Thus $f'(u_0) = 0$, $f'(u) < 0$ on $(0, u_0)$, and $f'(u) > 0$ on $(u_0, \infty)$. Thus, $f(u) \ge f(u_0)$ for all $u > 0$.
It suffices to prove $f(u_0) \ge 0$ that is \begin{align} &\frac{2(t^2 - 1)}{3(t-1) + \sqrt{(t - 1)(13t - 5)}} + \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2}\\ &\quad - t\ln t - 3(t-1)\ln \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2(t^2 - 1)^{2/3}} \ge 0. \end{align} The proof is not difficult (Hint: Let $\frac{13t-5}{t-1} = w^2$.). We are done.
$\phantom{2}$
Proof of Fact 2: Denote LHS by $g(u)$. We have \begin{align} g'(u) &= - (t^2-1)^{1/3}u^{-2} + (t^2 - 1)^{2/3} + 3(t-1)u^{-1}\\ &= \frac{1}{u^2}(u - u_1)\Big[(t^2 - 1)^{2/3} u + \frac{3(t-1) + \sqrt{(t - 1)(13t - 5)}}{2}\Big] \end{align} where $u_1 = \frac{2(t^2 - 1)^{1/3}}{3(t-1) + \sqrt{(t - 1)(13t - 5)}}$.
Thus $g'(u_1) = 0$, $g'(u) < 0$ on $(0, u_1)$, and $g'(u) > 0$ on $(u_1, \infty)$. Thus, $g(u) \ge g(u_1)$ for all $u > 0$.
It suffices to prove $g(u_1) \ge 0$ that is \begin{align} & \frac{3(t-1) + \sqrt{(t-1)(13t-5)}}{2} + \frac{2(t^2 - 1)}{3(t-1) + \sqrt{(t-1)(13t-5)}}\\ &\quad + t\ln t + 3(t-1)\ln \frac{2(t^2 - 1)^{1/3}}{3(t-1) + \sqrt{(t - 1)(13t - 5)}} \ge 0. \end{align} The proof is not difficult (Hint: Let $\frac{13t-5}{t-1} = w^2$.). We are done.