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How to solve these inequalities?

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

Any hints/solution are welcome.

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The first one looks like an application of Rearrangement inequality, assuming the partial order of $a,b,c,d$ from largest to smallest; second one looks like Jensen using $\varphi(x) = \frac{1}{x}$, but not positive? –  Shuhao Cao Aug 10 '11 at 16:47
    
I hadn't given it much thought, but maybe you can prove that $2^{n-1}(x_1\cdot\ldots\cdot x_n)>(x_1+1)\cdot\ldots\cdot(x_n+1)$ for all $n>1$ and $x_1,...,x_n>1$, by induction. This is clearly true for $n=2$. The case $n=4$ gives you 1. –  Dennis Gulko Aug 10 '11 at 17:00
    
I think second inequality has to have some sort of sign requirement, otherwise it is false, if both positive just apply Jensen's equality using $\varphi(\cdot) = \frac{1}{\cdot}$, $x_1 = 1/x$, $x_2 = 1/y$, $a_1 = axy$, $a_2 = bxy$ –  Shuhao Cao Aug 10 '11 at 17:13
    
@MathChief:I didn't knew about Jensen's inequality,but I believe there might be something more easy way?! –  Quixotic Aug 10 '11 at 17:47
    
For 2) to hold, assume that $x,y > 0$ and $x \neq y$. After some algebra, the inequality becomes $2xy < x^2 + y^2$... –  Shai Covo Aug 10 '11 at 17:49

1 Answer 1

up vote 6 down vote accepted

Solution:

  1. 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 $8(abcd + 1)> (a+1)(b+1)(c+1)(d+1)$.

  2. 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: $$ \frac{(a+b)xy}{ay+bx} = \varphi\left(\frac{axy\cdot \frac{1}{x} + bxy\cdot \frac{1}{y}}{axy + bxy}\right)<\frac{axy \cdot \varphi(\frac{1}{x})+bxy\cdot \varphi(\frac{1}{x})}{axy + bxy}= \frac{ax+by}{a+b}. $$

  3. 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, $x = \sqrt{3/2}, y= \sqrt{5/2}, z= \sqrt{7/2}$.

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Third problem can be done by choosing a correct weight and applying AMGM. –  Soarer Aug 10 '11 at 18:39
    
@Soarer: Thank you, done. –  Shuhao Cao Aug 10 '11 at 19:12

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