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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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    It is easy, starting from (axy + bxy)(a + b) < (ax + by)(ay + bx).2011-08-10

2 Answers 2

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

  • 0
    @Soarer: Thank you, done.2011-08-10
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Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=1+t$. Hence, $8(abcd + 1)-(a+1)(b+1)(c+1)(d+1)=$ $=8(1+x)(1+y)(1+z)(1+t)+8-(2+x)(2+y)(2+z)(2+t)=$ $=4(xy+xz+yz+xt+yt+zt)+6(xyz+xyt+xzt+yzt)+7xyzt>0$ Done!