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# 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 $$ab + bc + cd + da$$, subject to the conditions $$abcd = 4$$ and $$a^2 + b^2 + c^2 + d^2 =10$$.

Details and Assumptions :

• $$a$$, $$b$$, $$c$$ and $$d$$ 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
10 months, 2 weeks ago

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

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

Now, using the AM-GM inequality, we have

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

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

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

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

Therefore,

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

Let's consider two cases:

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

In this case, we have

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

Therefore,

\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 $$m=2n,\, n=\frac{5}{2}, a^2+b^2+c^2+d^2=10$$ and $$abcd=4$$. This can be attained for example when $$a=c=\sqrt{\frac{5}{2}},\quad b=\frac{\sqrt{205}+3\sqrt{5}}{10}, \quad d=\frac{\sqrt{205}-3\sqrt{5}}{10}$$.

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

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

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

And thus, it follows that

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

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

Proof No.2 (Way shorter)

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

$(a+c)(b+d)=\sqrt{(a^2+c^2+2ac)(b^2+d^2+2bd)}$

$= \sqrt{(a^2+c^2)(b^2+d^2)+2ac(b^2+d^2)+2bd(a^2+c^2)+4abcd}$

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

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

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

Solving, we have

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

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

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

$(a+c)(b+d) \leq \sqrt{82}$ · 10 months, 2 weeks ago

You made it too much lengthy. · 10 months, 2 weeks ago

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

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

$(a+c)(b+d)=\sqrt{(a^2+c^2+2ac)(b^2+d^2+2bd)}$

$= \sqrt{(a^2+c^2)(b^2+d^2)+2ac(b^2+d^2)+2bd(a^2+c^2)+4abcd}$

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

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

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

Solving, we have

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

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

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

$(a+c)(b+d) \leq \sqrt{82}$ · 10 months, 2 weeks ago

Awesome way · 10 months, 2 weeks ago

@Surya Prakash , Is this a better proof? · 10 months, 2 weeks ago

Yes this was exactly the way I solved it. · 10 months, 2 weeks ago

So you have a shorter solution? Can you please share on Slack if you don't want to put it here? · 10 months, 2 weeks ago

@Surya Prakash , the difficulty spiked dramatically here. Perhaps a bit too much. But please, continue posting olympiad questions like this. · 10 months, 2 weeks ago

Yes I also got √82 and it's correct. · 10 months, 2 weeks ago

Comment deleted 10 months ago

No · 10 months, 2 weeks ago

Comment deleted 10 months ago

No. Try it again · 10 months, 2 weeks ago

Is it √82? · 10 months, 2 weeks ago

I hate using latex, how can I attach an image? · 10 months, 2 weeks ago

Using Cauchy Scharw Inequality,

$(a^2 +b^2 + c^2 + d^2)(b^2 + c^2 + d^2 + a^2) \geq (ab+bc+cd+da)^2$

So $$10\geq (ab+bc+cd+da)$$. · 10 months, 2 weeks ago

Using this we get that equality holds when $$\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d} = \dfrac{d}{a}$$. Which gives us that$$a=b=c=d$$. So, from the equation $$a^2 + b^2 + c^2 + d^2 =10$$, we get that $$a=b=c=d=\sqrt{\dfrac{5}{2}}$$. This gives us that $$abcd=\dfrac{25}{4}$$. But $$abcd=4$$. So your solution is wrong. · 10 months, 2 weeks ago

OH! didnt notice that · 10 months, 2 weeks ago