Tiggsy

If there's a typo, and that 30 should be 31, then 4, 9, 25 would work.

Here's a slightly different parts routine

Let

\(\displaystyle I_{n}=\int1.\ln(x)^{n}dx\\ \text{and let } u=\ln(x)^{n}, \quad dv = 1.dx \)

Then

\(\displaystyle \frac{du}{dx}=n.\ln(x)^{n-1}\frac{1}{x},\quad v = x\)

so

\(\displaystyle I_{n}=x\ln(x)^{n}-\int xn\ln(x)^{n-1}\frac{1}{x}dx \\ I_{n}=x\ln(x)^n-nI_{n-1}.\)

The red and blue triangles are similar, so since their areas are in the ratio 27/12 = 9/4, their side lengths will be in the ratio 3/2.

So, let AB = 3k, then EF will equal 2k, (and notice in passing that DE + FC = 3k - 2k = k).

Run a perpendicular from AB to CD, passing through G and meeting the two sides at P and Q.

Let PG = 3h, then GQ = 2h.

Area of the red triangle = 3k*3h/2 = 27, so hk = 6.

(Similarly, area of the blue triangle = 2k*2h/2 = 12, so hk = 6).

Area of the trapezium ABCF = 5h*(AB + CF)/2  = 27 +Y1.

Area of the trapezium ABDE = 5h*(AB + DE)/2 = 27 + Y2.

Adding, 5h*(2AB + CF + DE)/2 = 54 + Y1 + Y2,    (DE + CF = k from earlier.)

so, 54 + Y1 + Y2 = 5h*(6k + k)/2 = 35hk/2 = 105,

so Y1 + Y2 = 51.

Drop a vertical from A meeting DC at E.

DE = (2y - 2x)/2 = y-x.

AD = x + y.

AE = 2R.

Now Pythagoras, AD^2 = AE^2 + DE^2, etc..

\(\displaystyle \frac{x^{2}-19}{x^{3}-2x^{2}-5x+6} \equiv \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3} \\ \text{so} \\ x^{2}-19 \equiv A(x+2)(x-3)+B(x-3)(x-1)+C(x-1)(x+2).\)

\(\displaystyle \text{Let }x=1,\qquad -18=A(3)(-2), \qquad A=3. \\ \text{Let }x=-2, \quad \, -15=B(-5)(-3), \qquad B= -1.\\ \text{Let }x=3,\qquad-10=C(2)(5),\qquad C=-1.\)

\(\displaystyle \left\{\frac{x^{102}-1}{x^{3}-1}-\frac{x^{101}-x^{2}}{x^{3}-1}\right\}(x^{2}+x+1)=x^{101}+x+1, \\ \text{so} \\ A+B=2.\)

The two fractions inside the curly brackets are both the sums of GP's.

The first is the sum of

 \(1+x^{3}+x^{6}+ \dots +x^{99},\)

and the second the sum of

\(x^{2}+x^{5}+ \dots + x^{98}.\)

Can't say that I fully understand this question, but here's some algebra that might help.

Consider

\(\displaystyle (a-1)^{2}+(b-24)^{2}+(c-38)^{2} ,\)

(geater than or equal to zero),  for all real values of a, b and c.

Expanding each bracket and rearranging,

\(\displaystyle a^{2}+b^{2}+c^{2}-2(a+24b+38c)+1^{2}+24^{2}+38^{2} \geq0, \\ \text{so} \\ a^{2}+b^{2}+c^{2}+2021 \geq2(a+24b+38c), \\ \text{and since } \\ a+24b+38c \geq 2021, \\a^{2}+b^{2}+c^{2} \geq 2(2021)-2021=2021.\)

The other part, a + 16b + 42c, leads to exactly the same result, but how that's to be presented to answer the question, I don't know.

\(\displaystyle f(x)=2-\cos(x)+\sin^{2}(x) \\=2-\cos(x)+1-\cos^{2}(x) \\ = 3-\{\cos^{2}(x)+\cos(x)+1/4-1/4\} \\ = 3-\{(\cos(x)+1/2)^{12}-1/4\} \\ =13/4-(\cos(x)+1/2)^{2}.\)

So, max will be 13/4 when cos(x) = -1/2, and min will be 13/4 - 9/4 = 1, when cos(x) = 1.

Treating the problem as a calculus max/min problem gets the result easily enough, but in keeping with the suggestion in the question here's the Cauchy-Schwarz solution.

For n = 3, Cauchy-Schwarz says that

\(\displaystyle a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \leq(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})^{1/2}(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})^{1/2},\)

so,

\(\displaystyle 1 =a+b+c=a.1+(\sqrt{2}b)(1/\sqrt{2})+c.1\leq(a^{2}+2b^{2}+c^{2})^{1/2}(1+\frac{1}{2}+1)^{1/2},\\ 1\leq(a^{2}+2b^{2}+c^{2})^{1/2}(5/2)^{1/2},\)

so squaring and cross-multiplying,

\(\displaystyle a^{2}+2b^{2}+c^{2}\geq2/5.\)

Thanks, much appreciated.