Waste less time on Facebook — follow Brilliant.
×

Cauchy-Schwarz Training

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.

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

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

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

Note by Guilherme Dela Corte
2 years, 11 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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

  1. Clearly \(2×5=5×2=10\).
  2. \(P (x)=(x-2)^{2}(x-5)(x-10)\)
  3. Using Cauchy-Schwarz rather than calculus is difficult, but here's a solution:

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, 11 months ago

Log in to reply

Can you explain adequately to your friends:

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, 11 months ago

Log in to reply

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.

Joel Tan - 2 years, 11 months ago

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, 11 months ago

Log in to reply

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.

Joel Tan - 2 years, 11 months ago

Log in to reply

Challenge: Can the third question be answered by Cauchy-Schwarz?

Guilherme Dela Corte - 2 years, 11 months ago

Log 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, 5 months ago

Log in to reply

Waht's the question which has no answer ?

Hazem Gamal - 2 years, 10 months ago

Log in to reply

All questions have answers! However, #1 is a proof, and thus has not a closed-form value to submit.

Guilherme Dela Corte - 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...