Online Course Discussion Forum

Q9 on the day 4 of week 1 of the camp

 
 
ZhangHenry的头像
Q9 on the day 4 of week 1 of the camp
ZhangHenry - 2020年07月18日 Saturday 17:03
 

Hi, I am reviewing the problems we did earlier and I'm a bit confused on some aspects of Q9 even with the recordings.

If we follow the solution presented in the video, how do we know we can let 2ab<=k1 *  a^2 + k2 *  b^2 where k1 = k2 + k3 = k4 + k5 = k6 and there will be some positive k's that satisfy this? Also once we generate those 3 inequalities that build to the thing we are trying to get with M in it, we also know that for all a and b and positive k1 and k2, 2sqrt(k1k2)ab <= k1a^2 + k2b^2. I get this expression and how we got it, but why does the sqrt(k1k2) now have to be 1 to match the above 2ab<=k1 *  a^2 + k2 *  b^2? Is this because they are both tight bounds so they must be equal?

thanks

 
WangDr. Kevin的头像
Re: Q9 on the day 4 of week 1 of the camp
WangDr. Kevin - 2020年07月19日 Sunday 14:54
 

The original question is the following:

(ZIML Varsity Feb 2018) Find the smallest $M$ such that the inequality \[ 2ab + 3bc + 2cd \leq M(a^2+b^2+c^2+d^2) \] holds for all groups of real numbers $a,b,c,d$.

Let me give a solution with more analysis. The inequality is symmetric in the variables $a$ and $d$ (so you can switch them without changing the inequality), and symmetric in $b$ and $c$.  For any real numbers $a$ ad $b$ and $k\neq 0$, the following inequality is true, based on AM-GM:

\[ ka^2 + \frac{1}{k}b^2 \geq 2ab. \]

Because of symmetry, we can have the following:

\[ kd^2 + \frac{1}{k}c^2 \geq 2cd. \]

And based on AM-GM, we also have:

\[ \frac{3}{2}b^2 + \frac{3}{2}c^2 \geq 3bc. \]

Adding these inequalities together, we get that the following is true for all $a,b,c,d,k$:

\[ ka^2 + \left(\frac{1}{k}+\frac{3}{2}\right)b^2 + \left(\frac{1}{k}+\frac{3}{2}\right)c^2 + kd^2 \geq 2ab + 3bc + 2cd. \]

So far there is only one parameter $k$. In order to make all the coefficients equal, we just need:

\[ k = \frac{1}{k} + \frac{3}{2}. \]

This has a positive solution $k=2$. Then, since the inequality holds for all $a,b,c,d$, and equality does hold when $b=c=2a=2d$, the value $M=2$ is the smallest possible value.

ZhangHenry的头像
Re: Q9 on the day 4 of week 1 of the camp
ZhangHenry - 2020年07月21日 Tuesday 14:35
 

Thanks, your solution is very clear. Since we need k = k + 3/2 for this to work and coefficients of the squares to be equal, does that mean that there is only one M that works for this question?

WangDr. Kevin的头像
Re: Q9 on the day 4 of week 1 of the camp
WangDr. Kevin - 2020年07月21日 Tuesday 18:43
 

Any number greater than $2$ will work.  This inequality has to be true for all $a, b, c, d$. If $M < 2$, there will be some values of $a,b,c,d$ so the inequality is false.  That means $M=2$ is the smallest possible value.

ZhangHenry的头像
Re: Q9 on the day 4 of week 1 of the camp
ZhangHenry - 2020年07月21日 Tuesday 19:53
 

Ok I got it, thanks!