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# 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

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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 $$x=\frac {64-4\sqrt {321}}{17-\sqrt {321}}$$.

- 2 years, 11 months ago

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

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

- 2 years, 11 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.

- 2 years, 11 months ago

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.).

- 2 years, 11 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.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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)

- 2 years, 5 months ago

Waht's the question which has no answer ?

- 2 years, 10 months ago

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

- 2 years, 10 months ago