Cauchy-Schwarz Training

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

The smallest real value of the expression ay2+by2 ay^2 + \dfrac{b}{y^2} can be written as cd c \cdot \sqrt{d} , where c c and dd are positive square-free integers. Restrict xx and y y to the real numbers.

  1. Show that ab=bc=dab = bc = d.

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

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

Note by Guilherme Dela Corte
4 years, 9 months ago

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The first expression can be reduced to sin2x+2cos2x+2\sin 2x+2\cos 2x+2.

(sin2x+2cos2x)2(sin22x+cos22x)(1+4)=5(\sin 2x+2\cos 2x)^{2} \leq (\sin^{2} 2x+ \cos^{2} 2x)(1+4)=5 hence the maximum value is 2+52+\sqrt {5}. Equality can occur when tanx=0.5\tan x=0.5.

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

(2y2+5y2)(5y2+2y2)(10+10)2=40(2y^{2}+\frac {5}{y^{2}})(\frac {5}{y^{2}}+2y^{2}) \geq (\sqrt {10}+\sqrt {10})^{2}=40 hence 2y2+5y240=2102y^{2}+\frac {5}{y^{2}} \geq \sqrt {40}=2\sqrt {10}.

Hence a=2,b=5,c=2,d=10a=2, b=5, c=2, d=10.

  1. Clearly 2×5=5×2=102×5=5×2=10.
  2. P(x)=(x2)2(x5)(x10)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, (a2+b2)(b2+a2)(ab+ba)2(a^{2}+b^{2})(b^{2}+a^{2}) \geq (ab+ba)^{2} hence a2+b22aba^{2}+b^{2} \geq 2ab for reals a,ba, b.

Thus for all positive reals a,b,c,da, b, c, d, a4+b4+c4+d42a2b2+2c2d24abcda^{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 (x5)(x10)(x-5)(x-10) negative, or 5<x<105 < x <10. Hence x2,x5,10xx-2, x-5, 10-x are positive. So an easier problem would be to find the maximum of (x2)(x2)(x5)(10x)(x-2)(x-2)(x-5)(10-x).

Now, using the fact above, r(x2)r(x2)(12r)(x5)(10x)(r(x2)+r(x2)+(12r)(x5)+10x4)4=(6r+54)4r(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 (6r+5)4256r2(12r)\frac {(6r+5)^{4}}{256r^{2}(1-2r)}.

In particular, when r=2132112r=\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=64432117321x=\frac {64-4\sqrt {321}}{17-\sqrt {321}}.

Joel Tan - 4 years, 9 months ago

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Can you explain adequately to your friends:

a) Why is the "maximum" (6r+5)4256r2(12r) \dfrac{(6r+5)^4}{256r^2 (1-2r)} ? How did attain it?

b) Why is r=2132112 r = \dfrac{21 - \sqrt{321}}{12} ?

Guilherme Dela Corte - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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

Guilherme Dela Corte - 4 years, 9 months ago

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Waht's the question which has no answer ?

Hazem Gamal - 4 years, 9 months ago

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

Guilherme Dela Corte - 4 years, 9 months ago

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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 - 4 years, 3 months ago

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