Olympiad Proof Problem - Day 3

This is the third proof problem of the series "Olympiad Proof Problems". Try this problem and post your working.


Find the maximum value of the expression ab+bc+cd+daab + bc + cd + da, subject to the conditions abcd=4abcd = 4 and a2+b2+c2+d2=10a^2 + b^2 + c^2 + d^2 =10.

Details and Assumptions :

  • aa, bb, cc and dd are positive real numbers

Try more proof problems at Olympiad Proof Problems.


EDIT: When dealing with inequalities, find out the condition of equality and check whether the equality condition satisfies the given constraints.

Note by Surya Prakash
3 years, 10 months ago

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I have a fairly long proof for this but hopefully, it is correct. Note that ab+bc+cd+da=(a+c)(b+d)ab+bc+cd+da=(a+c)(b+d).

WLOG, we may assume that acbdac \geq bd. From this, we have a2c2abcd=4a^2c^2 \geq abcd=4 and hence ac2ac \geq 2.

Now, using the AM-GM inequality, we have

10=(a2+c2)+(b2+d2)2ac+2bd=2ac+8ac.10=(a^2+c^2)+(b^2+d^2) \geq 2ac+2bd=2ac+\frac{8}{ac}.

Solving the inequality 2ac+8ac102ac+\frac{8}{ac} \leq 10, we get 1ac4.1 \geq ac \leq 4. Combining with the result ac2ac \geq 2 obtained above, we get 2ac42 \leq ac \leq 4.

Setting m=a2+c2m=a^2+c^2 and n=acn=ac (m2n,2n4m \geq 2n,\, 2 \leq n \leq 4). Under this substitution, we have

(a+c)2=m+2n,(b+d)2=b2+d2+2bd=10m+8n.(a+c)^2=m+2n,\quad (b+d)^2=b^2+d^2+2bd=10-m+\frac{8}{n}.

Therefore,

(a+c)2(b+d)2=(m+2n)(10m+8n)=(n+5+4n)2(m+n54n)2.(a+c)^2(b+d)^2=(m+2n)\left( 10-m+\frac{8}{n}\right) =\left(n+5+\frac{4}{n}\right)^2 -\left(m+n-5-\frac{4}{n}\right)^2.

Let's consider two cases:

Case 1: 3n25n+403n^2-5n+4 \geq 0

In this case, we have

m+n54n3n54n=3n25n4n0.m+n-5-\frac{4}{n} \geq 3n-5-\frac{4}{n}=\frac{3n^2-5n-4}{n} \geq 0.

Therefore,

(a+c)2(b+d)2(n+5+4n)2(3n54n)2=8[n(5n)+4]8{[n+(5n)2]2+4}=82.\begin{aligned} (a+c)^2(b+d)^2 & \leq \left(n+5+\frac{4}{n}\right)^2 -\left(3n-5-\frac{4}{n}\right)^2=8\left[n(5-n)+4\right] \\ &\leq 8 \left\{ \left[ \frac{n+(5-n)}{2}\right]^2+4\right\}=82.\end{aligned}

Equality holds when m=2n,n=52,a2+b2+c2+d2=10m=2n,\, n=\frac{5}{2}, a^2+b^2+c^2+d^2=10 and abcd=4abcd=4. This can be attained for example when a=c=52,b=205+3510,d=2053510a=c=\sqrt{\frac{5}{2}},\quad b=\frac{\sqrt{205}+3\sqrt{5}}{10}, \quad d=\frac{\sqrt{205}-3\sqrt{5}}{10}.

Case 2: 3n25n+4<03n^2 - 5n + 4 < 0

In this case, we have 2n<5+7362 \leq n < \frac{5+\sqrt{73}}{6}. Since the function f(n)=n+4n+5f(n)=n+\frac{4}{n}+5 is increasing for n2n \ge 2, we get

f(n)<f(5+736)=10+2733.f(n) <f\left(\frac{5+\sqrt{73}}{6}\right)=\frac{10+2\sqrt{73}}{3}.

And thus, it follows that

(a+c)2(b+d)2(n+5+4n)2<(10+2733)2<82.(a+c)^2(b+d)^2 \leq \left(n+5+\frac{4}{n}\right)^2 <\left(\frac{10+2\sqrt{73}}{3}\right)^2<82.

From this, it follows that the maximum value for (a+c)(b+d)=82.(a+c)(b+d)=\sqrt{82}.


Proof No.2 (Way shorter)

Once again, ab+bc+cd+da=(a+c)(b+d)ab+bc+cd+da=(a+c)(b+d). We have

(a+c)(b+d)=(a2+c2+2ac)(b2+d2+2bd)(a+c)(b+d)=\sqrt{(a^2+c^2+2ac)(b^2+d^2+2bd)}

=(a2+c2)(b2+d2)+2ac(b2+d2)+2bd(a2+c2)+4abcd= \sqrt{(a^2+c^2)(b^2+d^2)+2ac(b^2+d^2)+2bd(a^2+c^2)+4abcd}

(a2+b2+c2+d22)2+2(ab+cd)(ad+bc)+16\leq \sqrt{\left(\frac{a^2+b^2+c^2+d^2}{2}\right)^2 +2(ab+cd)(ad+bc)+16}

25+2(ab+bc+cd+da2)2+16\leq \sqrt{25+2\left(\frac{ab+bc+cd+da}{2}\right)^2+16}

=41+12(a+c)2(b+d)2=\sqrt{41+\frac{1}{2} (a+c)^2 (b+d)^2}

Solving, we have

(a+c)(b+d)41+12(a+c)2(b+d)2(a+c)(b+d) \leq \sqrt{41+\frac{1}{2} (a+c)^2 (b+d)^2}

(a+c)2(b+d)241+12(a+c)2(b+d)2(a+c)^2 (b+d)^2 \leq 41+\frac{1}{2} (a+c)^2 (b+d)^2

12(a+c)2(b+d)241\frac{1}{2} (a+c)^2 (b+d)^2 \leq 41

(a+c)(b+d)82(a+c)(b+d) \leq \sqrt{82}

Sharky Kesa - 3 years, 10 months ago

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@Surya Prakash , the difficulty spiked dramatically here. Perhaps a bit too much. But please, continue posting olympiad questions like this.

Sharky Kesa - 3 years, 10 months ago

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You made it too much lengthy.

Surya Prakash - 3 years, 10 months ago

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I do believe I have another proof which is much shorter. I can't believe I didn't think about this.

Once again, ab+bc+cd+da=(a+c)(b+d)ab+bc+cd+da=(a+c)(b+d). We have

(a+c)(b+d)=(a2+c2+2ac)(b2+d2+2bd)(a+c)(b+d)=\sqrt{(a^2+c^2+2ac)(b^2+d^2+2bd)}

=(a2+c2)(b2+d2)+2ac(b2+d2)+2bd(a2+c2)+4abcd= \sqrt{(a^2+c^2)(b^2+d^2)+2ac(b^2+d^2)+2bd(a^2+c^2)+4abcd}

(a2+b2+c2+d22)2+2(ab+cd)(ad+bc)+16\leq \sqrt{\left(\frac{a^2+b^2+c^2+d^2}{2}\right)^2 +2(ab+cd)(ad+bc)+16}

25+2(ab+bc+cd+da2)2+16\leq \sqrt{25+2\left(\frac{ab+bc+cd+da}{2}\right)^2+16}

=41+12(a+c)2(b+d)2=\sqrt{41+\frac{1}{2} (a+c)^2 (b+d)^2}

Solving, we have

(a+c)(b+d)41+12(a+c)2(b+d)2(a+c)(b+d) \leq \sqrt{41+\frac{1}{2} (a+c)^2 (b+d)^2}

(a+c)2(b+d)241+12(a+c)2(b+d)2(a+c)^2 (b+d)^2 \leq 41+\frac{1}{2} (a+c)^2 (b+d)^2

12(a+c)2(b+d)241\frac{1}{2} (a+c)^2 (b+d)^2 \leq 41

(a+c)(b+d)82(a+c)(b+d) \leq \sqrt{82}

Sharky Kesa - 3 years, 10 months ago

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@Sharky Kesa @Surya Prakash , Is this a better proof?

Sharky Kesa - 3 years, 10 months ago

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@Sharky Kesa Awesome way

Surya Prakash - 3 years, 10 months ago

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@Sharky Kesa Yes this was exactly the way I solved it.

Kushagra Sahni - 3 years, 10 months ago

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So you have a shorter solution? Can you please share on Slack if you don't want to put it here?

Sharky Kesa - 3 years, 10 months ago

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Yes I also got √82 and it's correct.

Kushagra Sahni - 3 years, 10 months ago

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Using Cauchy Scharw Inequality,

(a2+b2+c2+d2)(b2+c2+d2+a2)(ab+bc+cd+da)2(a^2 +b^2 + c^2 + d^2)(b^2 + c^2 + d^2 + a^2) \geq (ab+bc+cd+da)^2

So 10(ab+bc+cd+da)10\geq (ab+bc+cd+da).

Dev Sharma - 3 years, 10 months ago

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Using this we get that equality holds when ab=bc=cd=da\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = \dfrac{d}{a}. Which gives us thata=b=c=da=b=c=d. So, from the equation a2+b2+c2+d2=10a^2 + b^2 + c^2 + d^2 =10, we get that a=b=c=d=52a=b=c=d=\sqrt{\dfrac{5}{2}}. This gives us that abcd=254abcd=\dfrac{25}{4}. But abcd=4abcd=4. So your solution is wrong.

Surya Prakash - 3 years, 10 months ago

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OH! didnt notice that

Dev Sharma - 3 years, 10 months ago

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You only used the first condition that a^2+b^2+c^2+d^3=10. What about abcd=4?

Kushagra Sahni - 3 years, 10 months ago

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