Reals $a,b,c,x,y$ satisfies $a^2+b^2+c^2+x^2+y^2=1$. Find the maximum value of $$(ax+by)^2+(bx+cy)^2$$
Problem
Source: 2015 Korean Mathematical Olympiad P3
Tags: inequalities
rkm0959
01.11.2015 17:39
The maximum is $\frac{3}{8}$. The solution uses AM-GM only. (That's what I recall from the semi-official solution)
Sadly, I didn't guess the equality case right and did a little bit of BS for my "solution".
(I'm glad that I got the maximum value and the equality case though. Partial Credits )
sqing
02.11.2015 05:28
rkm0959 wrote: Reals $a,b,c,x,y$ satisfies $a^2+b^2+c^2+x^2+y^2=1$. Find the maximum value of $$(ax+by)^2+(bx+cy)^2$$ Reals $a,b,c,x,y$ satisfies $a^2+b^2+c^2=x^2+y^2=1$. Find the maximum value of $$(ax+by)^2+(bx+cy)^2$$http://www.artofproblemsolving.com/community/c6h1158074p5501289
mssmath
02.11.2015 14:56
Using Lagrange Multipliers I was able to solve this problem in less than half an hour. Again what is the point of giving such problem I ask?
rkm0959
02.11.2015 15:01
mssmath wrote: Using Lagrange Multipliers I was able to solve this problem in less than half an hour. Again what is the point of giving such problem I ask? Well, there is a beautiful solution with AM-GM. (I really need to learn this method. Can you post your solution?) Yeah I think so too. Algebra in KMO makes me sad a bit This year was LM Inequality/2002 IMO SL A4 variant. Last year was 5 minute FE/UVW Ineq. I would like to quote a student whom I don't know personally. "I got so much benefits from not trying problem 3"
mssmath
02.11.2015 15:05
Can you please post the solution by AM-GM?
rkm0959
02.11.2015 15:13
Alright here is the solution that made me groan and scream
This solution is not mine. It is the semi-official solution from saebon, which is a private institute.
Expanding, we have $a^2x^2+b^2(x^2+y^2)+c^2y^2+2xy(ab+bc)$.
Now by AM-GM, we have $a^2x^2+c^2y^2 = (a^2+c^2)(x^2+y^2)-(a^2y^2+c^2x^2) \le (a^2+c^2)(x^2+y^2)-2acxy$.
Also, we have $2xy(ab+bc)=2xyb(a+c) \le 2xy \cdot \left( \frac{b^2+(a+c)^2}{2} \right) = xy(a^2+b^2+c^2+2ac)=xy(a^2+b^2+c^2)+2acxy$.
Summing them all, we have $a^2x^2+b^2(x^2+y^2)+c^2y^2+2xy(ab+bc) \le (a^2+c^2)(x^2+y^2)+b^2(x^2+y^2)+xy(a^2+b^2+c^2)$
$\le (a^2+b^2+c^2)(x^2+y^2)+(\frac{x^2+y^2}{2})(a^2+b^2+c^2) =\frac{3}{2}(a^2+b^2+c^2)(x^2+y^2) \le \frac{3}{2}\cdot \left(\frac{a^2+b^2+c^2+x^2+y^2}{2}\right)^2=\frac{3}{8}$.
Equality holds when $a=c=\frac{1}{2\sqrt{3}}, b=\frac{1}{\sqrt{3}}, x=y=\frac{1}{2}$.
We don't need to find all the equality cases, so we are done. $\blacksquare$
Unfortunately, I guess the equality case wrong in the beginning -
I thought it would be $b^2=2a^2=2c^2=2x^2=2y^2$ for some reason.
I did some BS that <$x=y$ or $x=0$ or $y=0$> and <$a=c$ or $a=0$ or $c=0$> and then bashed.
At the final calculation when I used some discriminant to find the maximum, I realized that $b=2a=2c=\frac{2}{\sqrt{3}}x=\frac{2}{\sqrt{3}}y$. But 10 minutes were left. GG
There was a solution using Cauchy-Schwarz and "Greedy Algorithm" (I guess) posted on the local forum. Try to find it! I personally like that solution much more.
rightways
03.11.2015 04:05
rkm0959 wrote: Reals $a,b,c,x,y$ satisfies $a^2+b^2+c^2+x^2+y^2=1$. Find the maximum value of $$(ax+by)^2+(bx+cy)^2$$ how much time total was given for first day Senior KMO with 4 problems ?
sqing
03.11.2015 05:05
rkm0959 wrote: Well, there is a beautiful solution with AM-GM. (I really need to learn this method. Can you post your solution?) Yeah I think so too. Algebra in KMO makes me sad a bit This year was LM Inequality/2002 IMO SL A4 variant. Last year was 5 minute FE/UVW Ineq. I would like to quote a student whom I don't know personally. "I got so much benefits from not trying problem 3" Do not understand： This year was LM Inequality/2002 IMO SL A4 variant. Last year was 5 minute FE/UVW Ineq. Thanks.
rkm0959
03.11.2015 06:57
2015 KMO P3 (this problem) can be solved with Lagrange Multipliers. 2015 KMO P5 is a functional equation that is a variant of 2002 IMO Shortlist A4, as pointed out by trumpeter. 2014 KMO P2 was a very easy functional equation. 2014 KMO P7 was an inequality that can be solved very easily if you know the UVW-theorem.
sqing
03.11.2015 11:22
Thank you very much.
mssmath
04.11.2015 02:46
Much nicer solution
Notice that \[(ax+by)^2+(bx+cy)^2\le(ax+by)^2+(bx+cy)^2+(cx-ay)^2\]\[=(x^2+y^2)(a^2+b^2+c^2)+(2ab+2bc-2ab)xy\le\frac{3}{2}(x^2+y^2)(a^2+b^2+c^2)\le\frac{3}{8}\]The second to last inequality follows as $2ab+2bc-2ab\le a^2+b^2+c^2$ which is equivalent to $(a+c-b)^2\ge0$.
Vexation
18.12.2015 18:53
\[ (ax+by)^2 + (bx+cy)^2 = b^2 (x^2 +y^2 ) + a^2 x^2 + 2bxy(a+c) + c^2 y^2 \]By subtituting $x=r\sin\theta$ and $y=r\cos\theta$ where $r^2 = 1-a^2 - b^2-c^2$, we get \[ a^2 x^2 + 2bxy(a+c) + c^2 y^2 = r^2 (a^2 \sin^2 \theta + c^2 \cos^2 \theta + 2b(a+c)\sin \theta \cos\theta)\], whose maximum is $r^2 \{ \frac {a^2 + c^2 }{ 2} + \sqrt{b^2 (a+c)^2 + ( \frac{c^2 - a^2}{2})^2 }\}$. Therefore, \[b^2 (x^2 +y^2 ) + a^2 x^2 + 2bxy(a+c) + c^2 y^2 \le (1-a^2 - b^2 - c^2 ) (b^2 + \frac {a^2 + c^2 }{ 2} + \sqrt{b^2 (a+c)^2 + ( \frac{c^2 - a^2}{2})^2 }).\]Let $f(a,b,c) = b^2 + \frac {a^2 + c^2 }{ 2} + \sqrt{b^2 (a+c)^2 + ( \frac{c^2 - a^2}{2})^2 } $. Both $a^2 + b^2 + c^2$ and $f(a,b,c)$ are homogenous of degree $2$, so we can assume $a^2 + b^2 + c^2 = 1/2$. We rewrite $f(a,b,c)$ in terms of $b$ and $ac$ as following; \[ g(ac,b) = \frac{1}{4} + \frac{b^2}{2} + \sqrt{\frac{1}{16} + \frac{b^2}{4} - \frac{3b^4}{4} + 2(ac)b^2 - (ac)^2}\]1) $b^2 \le \frac{1}{6}$ $g(ac,b)$ is maximized when $ac=b^2$, which is possible for $b^2 = ac \le \frac{a^2+c^2}{2} = \frac{1}{4} - \frac{b^2}{2}$. \[g(b^2,b)= \frac{1}{4} + \frac{b^2}{2} + \sqrt{(\frac{1}{4} + \frac{b^2 }{2})^2} = b^2 + \frac{1}{2} \le \frac{2}{3}.\]2) $b^2 \ge \frac{1}{6}$ Since $ac$ cannot reach $b^2$, $a=c$ is the best possible for its maximum. $d = \frac{a^2+c^2}{2}$. \[f(a,b,c) \le f(d,b,d) = \frac{1}{4} + \frac{b^2}{2} + b\sqrt{1-2b^2}\], whose maximum is found at $b=\frac{1}{\sqrt{3}}$ as $\frac{3}{4}$.
seoneo
05.07.2017 14:43
Freeze $x$, $y$ and view as a quadratic form of the variables $a$, $b$, $c$. The eigenvalues of the corresponding matrix is $0$, $x^2+y^2 \pm xy$. Since $a^2+b^2+c^2 = 1-x^2-y^2$, we have the maximum \[ (1-x^2-y^2)(x^2+y^2 + |xy|) \]for each x,y such that $x^2 + y^2 \le 1$. Let $t=x^2+y^2$. Then $0 \le t \le 1$ and \[ (1-x^2-y^2)(x^2+y^2 + |xy|) \le \frac{3}{2} (1-t)t \le \frac{3}{8} \]Equality holds when, $|x|=|y|= 1/2$ and $(a,b,c)$ is the eigenvector for the corresponding eigenvalue.