Problem of today's MATH190 tutorial lesson

Problems typed below can also be found in this tutorial note, http://ihome.ust.hk/~masoarer/math190/week3.pdf. For fear that the web host will be out of service due to any reason, I have typed everything below. This text is prepared for future use, for example, my ``collection of problems".

Exercise 1. Let $  a,b,c>0$, show that $  (a^3+1)(b^3+1)(c^3+1)\ge (a^2b+1)(b^2c+1)(c^2a+1)$.

My way.
Note that $  (a^3+1)(a^3+1)(b^3+1)\ge (a^2b+1)^3$, we are done (you can see what happens when $  (a,b)=(b,c)$ and $  (a,b)=(c,a)$.).

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Exercise 2. If $  a,b,c>0$, prove that $  \displaystyle \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3}\leq \sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}$.

My way.
We see that \begin{align*}
 a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}&=\left(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\right)\left(\frac{a}{3}+\frac{a+b}{6}+\frac{b}{3}\right)\left(\frac{a}{3}+\frac{b}{3}+\frac{c}{3}\right)\\
&\ge \left(\frac{a}{3}+\sqrt[3]{\frac{ab(a+b)}{2\cdot 3^3}}+\frac{\sqrt[3]{abc}}{3}\right)^3.
\end{align*} Finally, $  \displaystyle\frac{a+b}{2}\ge\sqrt{ab}$ does solve the problem.

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Exercise 5. Let $  a,b,c>0$, prove that $  \displaystyle \frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2}\ge \frac{abc(a+b+c)}{2}$.

My way.
Use exercise 4 once, we have a new lower bound $  \displaystyle \frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2}\ge \frac{1}{6}(a^2+b^2+c^2)^2$, since \[(a^2+b^2+c^2)^2 \ge (ab+bc+ca)^2\ge 3(abbc + bcca + caab)=3abc(a+b+c),
\] we are done.

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Exercise 6. Let $  f$ be convex on $  [a,b]$. If $  c,d\in[a,b]$ with $  c-a>b-d$, prove that $  \displaystyle 2f\left(\frac{c+d}{2}\right) \leq f(c)+f(d)\leq f(c+d-b)+f(b)$.

My way.
The left inequality is obviously true. To go through right inequality, we first determine the position of $  a,b,c,d$ and $  c+d-b$. Observe that the inequality will change nothing if we interchange $  c$ and $  d$, so without loss of generality, we assume that $  c\leq d$, finally $  c+d-b\leq c \iff d\leq b$ we have known that $  a\leq c+d-b\leq c\leq d\leq b$.

Since $  \displaystyle f(c)+f(d)\leq f(c+d-b)+f(b)\iff \frac{f(b)-f(c)}{b-c}\ge \frac{f(d)-f(c+d-b)}{d-(c+d-b)}$, the equivalent inequality is obvious by convexity.

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Exercise 9. Let $  A,B,C$ be angles of a triangle. Prove that $  \displaystyle \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\leq \frac{9\sqrt{3}}{2\pi}$.

My way.
We observe the ineqality $  \displaystyle \frac{\sin x}{x}\leq \left(\frac{\frac{\pi}{3}\cdot \frac{1}{2}-\frac{\sqrt{3}}{2}}{\left(\frac{\pi}{3}\right)^2}\right)\left(x-\frac{\pi}{3}\right)+\frac{3\sqrt{3}}{2\pi}$. I haven't rigorously checked its validity by calculus, at least it is true for $  x\in (0,\pi)$ with the aid of graph ploting, hence replacing $  x$ by respectively $  a,b,c$ and adding them up, we get desired result.

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Exercise 10. Let $  f$ be a convex function on $  [a,b]$ and $  x,y,z\in [a,b]$. Prove by majorization inequality that $  \displaystyle f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\ge 2 \left(f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right)\right)$.

My way.
Correcting.

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Exercise 3, 4, 7, 8 are skipped (I would not include these problems in my pdf), I am trying exercise 10.