The **biggest** real value of the expression \( \cos{x} \cdot ( 2\sin{x} + 4\cos{x})\) can be written as \( a+ \sqrt{b} \), where \( a \) and \(b \) are positive square-free integers.

The **smallest** real value of the expression \( ay^2 + \dfrac{b}{y^2} \) can be written as \( c \cdot \sqrt{d} \), where \( c \) and \(d \) are positive square-free integers. Restrict \(x \) and \( y \) to the real numbers.

Show that \(ab = bc = d\).

Find the monic polynomial \(P(x) \) of fourth degree that has \( a,b,c,d \) as its roots.

Find \( P(x) \) 's smallest real value.

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TopNewestThe first expression can be reduced to \(\sin 2x+2\cos 2x+2\).

\((\sin 2x+2\cos 2x)^{2} \leq (\sin^{2} 2x+ \cos^{2} 2x)(1+4)=5\) hence the maximum value is \(2+\sqrt {5}\). Equality can occur when \(\tan x=0.5\).

Now \(c, d\) can be found by AM-GM. However, since Cauchy-Schwarz needs to be used, we do it in a slightly different way.

\((2y^{2}+\frac {5}{y^{2}})(\frac {5}{y^{2}}+2y^{2}) \geq (\sqrt {10}+\sqrt {10})^{2}=40\) hence \(2y^{2}+\frac {5}{y^{2}} \geq \sqrt {40}=2\sqrt {10}\).

Hence \(a=2, b=5, c=2, d=10\).

By Cauchy-Schwarz, \((a^{2}+b^{2})(b^{2}+a^{2}) \geq (ab+ba)^{2}\) hence \(a^{2}+b^{2} \geq 2ab\) for reals \(a, b\).

Thus for all positive reals \(a, b, c, d\), \(a^{4}+b^{4}+c^{4}+d^{4} \geq 2a^{2} b^{2}+2c^{2} d^{2} \geq 4abcd\) by applying the above twice. We will use this fact.

Clearly the polynomial can have negative values. Just fit in x=7.

Now the only way to get negative values is to have \((x-5)(x-10)\) negative, or \(5 < x <10\). Hence \(x-2, x-5, 10-x\) are positive. So an easier problem would be to find the maximum of \((x-2)(x-2)(x-5)(10-x)\).

Now, using the fact above, \(r(x-2)r (x-2)(1-2r)(x-5)(10-x) \leq (\frac {r(x-2)+r (x-2)+(1-2r)(x-5)+10-x}{4})^{4}=(\frac {6r+5}{4})^{4}\). So the maximum, if equality can occur, is \(\frac {(6r+5)^{4}}{256r^{2}(1-2r)}\).

In particular, when \(r=\frac {21-\sqrt {321}}{12}\), we get the maximum as that is the only case equality can occur. (Remember that the corresponding terms in Cauchy-Schwarz have to be related by a common ratio for equality.) The minimum is the negative of the maximum and is approximately -223. Equality can occur at \(x=\frac {64-4\sqrt {321}}{17-\sqrt {321}}\). – Joel Tan · 2 years, 6 months ago

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a) Why is the "maximum" \( \dfrac{(6r+5)^4}{256r^2 (1-2r)} \) ? How did attain it?

b) Why is \( r = \dfrac{21 - \sqrt{321}}{12}\) ? – Guilherme Dela Corte · 2 years, 6 months ago

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– Joel Tan · 2 years, 6 months ago

a) it is by rearranging the terms in the inequality. b) By Cauchy, equality can occur only when r(x-2)=(1-2r)(x-5)=10-x. Solving gives r.Log in to reply

Can anyone tell me about Cauchy-Schwarz inequality? I know the AM-GM inequality but not this. Please illustrate some simple examples too (I am new to this inequality.).

Thanks in advance! – Ninad Akolekar · 2 years, 6 months ago

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– Joel Tan · 2 years, 6 months ago

If you type in Cauchy-Schwarz in the search box there would be a featured community post on Cauchy-Schwarz. There's a lot of information there.Log in to reply

Challenge: Can the third question be answered by Cauchy-Schwarz? – Guilherme Dela Corte · 2 years, 6 months agoLog in to reply

Write a comment or ask a question...Find K so that U and V are orthogonal where U=(2,3K,-4,1,-5) and V=(6,-1,3,7,2K) – Sunny Bassey · 2 years ago

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Waht's the question which has no answer ? – Hazem Gamal · 2 years, 6 months ago

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#1is a proof, and thus has not a closed-form value to submit. – Guilherme Dela Corte · 2 years, 6 months agoLog in to reply