How to solve these inequalities? How to solve these inequalities?

  
*
  
*If $a,b,c,d \gt 1$, prove that $8(abcd + 1) \gt
 (a+1)(b+1)(c+1)(d+1)$.
  
*Prove that $ \cfrac{(a+b)xy}{ay+bx} \lt \cfrac{ax+by}{a+b}$
  
*Find the greatest value of $x^3y^5z^7$ when $2x^2+2y^2+2x^2=15$
  

Any hints/solution are welcome.

Solution:


*

*Since $a,b,c,d>1$, then the following inequalities are true based on Rearrangement inequalities: if $x>1$ and $y>1$ then $(x-1)(y-1) > 0$, ie $xy+1 >x+y$.
$$
\begin{aligned}
abcd + 1 &> abc + d
\\
abcd + 1 &> abd + c
\\
abcd + 1 &> acd + b
\\
abcd + 1 &> bcd + a
\\
abcd + 1 &> ab + cd
\\
abcd + 1 &> ad + bc
\\
abcd + 1 &> ac + bd
\\
abcd + 1 &= abcd + 1
\end{aligned}
$$
Adding them all up you get $8(abcd + 1)> (a+1)(b+1)(c+1)(d+1)$.

*Assuming $a,b,x,y>0$, and $x\neq y$, using Jensen's inequality 
$$
\varphi(\frac{a_1 t_1+ a_2 t_2}{a_1 + a_2}) \leq \frac{a_1 \varphi(t_1)+a_2 \varphi(t_2)}{a_1 + a_2}
$$
where $\varphi$ is a convex function. Here take $\displaystyle \varphi(t) = \frac{1}{t}$, $\displaystyle t_1=\frac{1}{x}, t_2=\frac{1}{y}$, $a_1=axy, a_2=bxy$ , apply the inequality you have:
$$
\frac{(a+b)xy}{ay+bx} = \varphi\left(\frac{axy\cdot \frac{1}{x} + bxy\cdot \frac{1}{y}}{axy + bxy}\right)<\frac{axy \cdot \varphi(\frac{1}{x})+bxy\cdot \varphi(\frac{1}{x})}{axy + bxy}= \frac{ax+by}{a+b}.
$$

*Let $r^2 = 15/2$, then use spherical coordinates, or calculus. Or using AM-GM inequality, write 
$$
\begin{aligned}
x^3 y^5 z^7 &= \frac{1}{3^{5/2}\cdot 5^{3/2}\cdot (15/7)^{7/2}} \cdot \left((5x^2)^{1/5} \cdot (3y^2)^{1/3}\cdot (\frac{15}{7} z^2)^{7/15}\right)^{15/2}
\\
&\leq \frac{1}{3^{5/2}\cdot 5^{3/2}\cdot (15/7)^{7/2}} \cdot \left(\frac{1}{5}\cdot 5x^2 +  \frac{1}{3}\cdot 3y^2+ \frac{7}{15}\cdot \frac{15}{7} z^2\right)^{15/2}
\\
&=  \frac{1}{3^{5/2}\cdot 5^{3/2}\cdot (15/7)^{7/2}}\cdot (\frac{15}{2})^{\frac{15}{2}}
\end{aligned}
$$
the maximum is obtained at $5x^2 = 3y^2 = \frac{15}{7} z^2$, ie, $x = \sqrt{3/2}, y= \sqrt{5/2}, z= \sqrt{7/2}$.
