coba pakai mathjax

LAT 1, yang sulit no 8,9

1.	$x+ \frac{1}{x} > 2$ $x^2 + 1 > 2x$ $(x-1)^2 > 0$

2.	$\frac{x}{y} + \frac{y}{x} > 2$ $x^2 + y^2 > 2xy$ $(x-y)^2 > 0$

3.	$x+y = 1$, maka $ab \le \frac{1}{2}$ $\left( \sqrt{x}-\sqrt{y}\right) \ge 0$ $x+y - 2\sqrt{xy} \ge 0$ $1 \ge 2\sqrt{xy}$ $\frac{1}{4} \ge xy$ $xy \le \frac{1}{4}$

4.	$(pa + qb)(pq+ab) \ge 2\sqrt{paqb} .2\sqrt{pqab}$ $\ge 4 pqab$

5.

${b_1+...+b_8 \over 8} = {{{b_1 + b_2 \over 2}+{b_3 + b_4 \over 2} \over 2} +{{b_5 + b_6 \over 2}+{b_7 + b_8 \over 2} \over 2} \over 2}$ $\ge {{\sqrt{b_1 . b_2}+\sqrt{b_3 . b_4} \over 2} +{\sqrt{b_5 . b_6}+\sqrt{b_7 . b_8} \over 2} \over 2}$ $\ge {\sqrt[4]{b_1 . b_2 . b_3 . b_4}+\sqrt[4]{b_5 . b_6 . b_7 . b_8} \over 2}$ $\ge \sqrt[8]{b_1 . b_2 . b_3 . b_4 . b_5 . b_6 . b_7 . b_8}$

6.	${a^2 \over 1+a^4} \le {1 \over 2}$ $ 2a^2 \le 1 + a^4$ $ a^4 - 2a^2 +1 \ge 0$ $ (a^2 -1)^2 \ge 0$

7.	${a^2 + 2 \over \sqrt{a^2 +1} }\ge 2$ $a^2 + 1 + 1 \ge 2 \sqrt{a^2 +1}$ $\left( \sqrt{a^2 + 1}-1 \right)^2 \ge 0$

8.	Untuk setiap bilangan asli a,b dengan $a>b$, buktikan bahwa $a! + b! \ge (a-1)! + (b+1)!$ untuk bilangan asli, $a-b \ge 1$ $a -1 \ge b$ $(a-1)! \ge b!$ $(a-1)(a-1)! \ge b b!$ $a(a-1)! - (a-1)! \ge (b+1 -1)b!$ $a! - (a-1)! \ge (b+1)b! - b!$ $a! - (a-1)! \ge (b+1)! - b!$ $a! + b! \ge (a-1)! + (b+1)!$

9.

Untuk setiap Bilangan asli a,b dengan $a \ge 2 , b \ge 2, a \ne b$ dan $a+b = 2p$, dengan p bilangan asli, sehinggga $p \ge 3$, perlihatkan bahwa $a! + b! > 2p!$ misal $a>b , a=p+n , b=p-n , p-n \ge 1 $ untuk n = 1 (p+1)!=(p+1)p! >2p! untuk n=k (p+k)!= (p+k)(p+k-1)...(p+2)(p+1)p! > k(k-1)(k-2)...2.1.p! = k!p! (p+k)! > k!p! > 2p!

10.	$x^3 + y^3 > x^2 y + xy^2$ $(x+y)^3 - 3xy(x+y) > xy(x+y)$ $(x+y)^2 - 3xy > xy$ $(x-y)^2 > 0$

11.	a.	$(x+y+z)^2 > 3(xy + yz + zx)$
		$x^2 + y^2 +z^2 +2(xy + yz + zx)> 3(xy + yz + zx)$
		$x^2 + y^2 +z^2 > xy + yz + zx$
		${x^2 + y^2 \over 2} + {y^2 + z^2 \over 2} + {x^2 + z^2 \over 2}> xy + yz + zx$

b.

$ xyz(x+y+z) \\ = xyzx + xyzy + xzyz \\ \lt \frac{x^2y^2 + z^2x^2}{2} + \frac{x^2y^2 + z^2y^2}{2} + \frac{x^2z^2 + y^2z^2}{2} \\ \lt x^2y^2 + x^2z^2 + y^2z^2 \\ $ $ xyz(x+y+z) \lt x^2y^2 + x^2z^2 + y^2z^2 \\ 3xyz(x+y+z) \lt 2xyz(x+y+z) + x^2y^2 + x^2z^2 + y^2z^2 \\ 3xyz(x+y+z) \lt (xy + xz + yz)^2 $

12. Jika x,y,z panjang sisi-sisi segitiga XYZ dengan z sisi miring buktikan bahwa

	a.	$z^3 \gt x^3 + y^3$ $(x^2 + y^2)^{3 \over 2} \gt x^3 + y^3$ $(x^2 + y^2)^3 \gt (x^3 + y^3)^2$ $x^6 + y^6 + 3x^2 y^2 (x^2 + y^2) \gt x^6 + y^6 + 2x^3 y^3$ $x^6 + y^6 + 3x^2 y^2 (x^2 + y^2) \gt x^6 + y^6 + 3x^2 y^2 (\frac{2}{3} xy)$ tinggal membuktikan $(x^2 + y^2) \gt \frac{2}{3} xy$ $(x^2 + y^2) \ge 2xy \gt (\frac{2}{3} xy)$
	b.	$z^4 \gt x^4 + y^4$ $(x^2 + y^2)^2 \gt x^4 + y^4$ $x^4 + y^4 + 2x^2 y^2 \gt x^4 + y^4$
	c.	$z^n \gt x^n + y^n$ untuk n bilangan genap $(x^2 + y^2)^\frac{n}{2} \gt x^n + y^n$ $x^n + y^n + \binom{n \over 2}{1}x^{n-2}y^2 + \binom{n \over 2}{2}x^{n-4}y^4 + ... + \binom{n \over 2}{{n \over 2}-2}x^{4}y^{n-4} + \binom{n \over 2}{{n \over 2}-1}x^{2}y^{n-2} \gt x^n + y^n$ untuk n bilangan ganjil $(x^2 + y^2)^\frac{n}{2} \gt x^n + y^n$ $(x^2 + y^2)^n \gt (x^n + y^n)^2$ $x^{2n} + y^{2n} + \binom{n}{1} x^{2n-2} y^2 + \binom{n}{2} x^{2n-4} y^4 + ... + \binom{n}{\frac{n-1}{2}} x^{n+1} y^{n-1} + \binom{n}{\frac{n+1}{2}} x^{n-1} y^{n+1} + ... + \binom{n}{n-2} x^{4} y^{2n -4} + \binom{n}{n-1} x^{2} y^{2n-2} \gt x^{2n} + y^{2n} +2x^ny^n$ seperti 13.a $\binom{n}{\frac{n-1}{2}} = \binom{n}{\frac{n+1}{2}}$ $2 \binom{n}{\frac{n+1}{2}} x^{n-1}y^{n-1} (x^2 + y^2) \gt 2x^{n-1}y^{n-1}(xy)$ $2 \binom{n}{\frac{n+1}{2}} x^{n-1}y^{n-1} (x^2 + y^2) \gt 2 \binom{n}{\frac{n+1}{2}} x^{n-1}y^{n-1}\left({1\over \binom{n}{\frac{n+1}{2}} }xy\right)$ $(x^2 + y^2) \ge 2xy \gt \left({1\over \binom{n}{\frac{n+1}{2}} }xy\right)$