mrdoctorlee

Cho a b c >0 và a+b=c=1. Tìm min F=$\frac{9}{1-2(ab+ac+bc)}+\frac{2}{abc}$

ta cos $\frac{9}{1-2(ab+bc+ac)}=\frac{9}{(a+b+c)^{2}-2(ab+bc+ac)}=\frac{9}{a^{2}+b^{2}+c^{2}}$

conf $\frac{2}{abc}=\frac{2(a+b+c)}{abc}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\geq \frac{2.9}{ab+bc+ac}$

vt $\geq \frac{9}{a^{2}+b^{2}+c^{2}}+\frac{9}{ab+bc+ac}+\frac{9}{ab+bc+ac}\geq \frac{(3+3+3)^{2}}{(a+b+c)^{2}}=81$

dau =sayr ra khi a=b=c=1/3

Cho x y z>0 và $\frac{1}{x+1}+\frac{2}{y+2}+\frac{3}{z+3}=1$ Tìm min P=xyz

ta có $1-\frac{1}{x+1}=\frac{2}{y+2}+\frac{3}{z+3}\geq 2\sqrt{\frac{6}{(y+2)(z+3)}}$

hay $\frac{x}{x+1}\geq 2\sqrt{\frac{6}{(y+2)(z+3)}}$

ttự $\frac{y}{y+2}\geq 3\sqrt{\frac{3}{(x+1)(z+3)}}$

$\frac{z}{z+3}\geq 2\sqrt{\frac{2}{(x+1)(y+2)}}$

nhân vế vs vế các bdt cumgf cchiều ta có $\frac{xyz}{(x+1)(y+2)(z+3)}\geq 8\sqrt{\frac{6.3.2}{(x+1)^{2}(y+2)^{2}(z+3)^{2}}}$

suy ra xyz>=48

dau = xảy ra khii $\inline \frac{1}{x+1}=\frac{2}{y+2}=\frac{3}{z+3}$

hay x=2, y=4,z=6

Tìm max P=$\frac{2010x^2+6024x+6\sqrt{x^3-2x^2+x-2}-8027}{x^2+3x-4}$

P= $\frac{2010(x^{2}+3x-4)+6\sqrt{(x^{2}+1)(x-2)}-6x+13}{x^{2}+3x-4}$

$= 2010+\frac{2.\sqrt{(x^{2}+1)(9x-18)}-6x+13}{x^{2}+3x-4}$$\leq 2010+\frac{x^{2}+1+9x-18-6x+13}{x^{2}+3x-4}=2010+1=2011$

dau = xảy ra khi x^2+1=9x-18 hay x= $\frac{9+\sqrt{157}}{2}$(vì X>=2) theo dkxd)

đặt $\sqrt{2x+1}=a\Rightarrow a^{2}=2x+1\Rightarrow x=\frac{a^{2}-1}{2}$

ttự $b=\sqrt{3y+1}\Rightarrow y=\frac{b^{2}-1}{3}$

$\sqrt{4z+1}=c\Rightarrow z=\frac{c^{2}-1}{4}$

nên x+y+z=$\frac{a^{2}-1}{2}+\frac{b^{2}-1}{3}+\frac{c^{2}-1}{4}=4$

$\Leftrightarrow \frac{a^{2}}{2}+\frac{b^{2}}{3}+\frac{c^{2}}{4}=\frac{61}{12}\Rightarrow a^{2}+\frac{2}{3}b^{2}+\frac{1}{2}c^{2}=\frac{61}{6}$

* tìm max p =a+b+c

 ta có $a^{2}+\frac{61}{27}\geq 2\sqrt{\frac{61}{27}}a$

   $\frac{2}{3}(b^{2}+\frac{61}{12})\geq \frac{4}{3}\sqrt{\frac{61}{12}}b$

$\frac{1}{2}(c^{2}+\frac{244}{27})\geq \sqrt{\frac{244}{27}}c$

cộng vế vs vế các bdt cùng chiều

$a^{2}+\frac{2}{3}b^{2}+\frac{1}{2}c^{2}+\frac{61}{6}\geq \frac{2\sqrt{183}}{9}(a+b+c)$

hay P<= $\frac{\sqrt{183}}{2}$

đấu = xảy ra khi a=$\sqrt{\frac{61}{27}}\Rightarrow x=\frac{17}{27}$

ttụ y= $\frac{49}{36};z=\frac{217}{108}$

* tìm min P =a+b+c

P^2 =(a+b+c)^2 =$a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=(a^{2}+\frac{2}{3}b^{2}+\frac{1}{2}c^{2})+\frac{1}{3}b^{2}+\frac{1}{2}c^{2}+2(ab+bc+ac)=\frac{61}{6}+\frac{1}{3}b^{2}+\frac{1}{2}c^{2}+2(ab+bc+ac)$

mà $a=\sqrt{2x+1}\geq 1$ ttự b>=1 , c>=1

nên P^2 >= $\frac{61}{6}+\frac{1}{3}+\frac{1}{2}+2(ab+bc+ac)=11+2(ab+bc+ac)$

vì a,b,c>=1

nên ta có (a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)>=0 hay

ab+bc+ac>=2(a+b+c)-3

suy ra 2(ab+bc+ac)>=4(a+b+c)-6 =4P-6

neen P^2>=11+4P-6

$P^{2}-4P-5\geq 0$ nên hoặc P<=-1 hoặc P>=5 (1)

mà P=a+b+c>=3 (2)

từ (1)(2) ta suy ra P>=5

dau = say ra khi a=3, b=c=1 suy ra x=4 y=z=0

Giải PT

$4\sqrt{5x^{3}-6x^{2}+2}+4\sqrt{-10x^{3}+8x^{2}+7x-1}=13-x$

$4\sqrt{5x^{3}-6x^{2}+2} =4.1.\sqrt{5x^{3}-6x^{2}+2}\leq 2(5x^{3}-6x^{2}+2+1)=10x^{3}-12x^{2}+6$

$4\sqrt{-10x^{3}+8x^{2}+7x-1}=2.\sqrt{4}.\sqrt{-10x^{3}+8x^{2}+7x-1}\leq -10x^{3}+8x^{2}+7x-1+4=-10x^{3}+8x^{2}+7x+3$

13-x=vt$\leq 10x^{3}-12x^{2}+6-10x^{3}+8x^{2}+7x+3=-4x^{2}+7x+9$

hay

$-4x^{2}+8x-4\geq 0\Leftrightarrow -4(x-1)^{2}\geq 0$

điều này xảy ra khi x=1