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.

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## Comments

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TopNewestI 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

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

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

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

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

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

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You made it too much lengthy.

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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, \(ab+bc+cd+da=(a+c)(b+d)\). 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

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

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

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

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

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@Surya Prakash , Is this a better proof?

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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?

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Yes I also got √82 and it's correct.

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

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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.

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OH! didnt notice that

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

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