a) áp dụng BĐT cô-si ta có:

\(y=\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}.\frac{18}{x}}=2\sqrt{9}=6\)

Dấu "=" xảy ra khi:

\(\frac{x}{2}+\frac{18}{x}=6\)

\(\Leftrightarrow\frac{x^2}{2x}+\frac{36}{2x}=\frac{12x}{2x}\)

\(\Rightarrow x^2+36=12x\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x=6\)

tương tự mấy câu tiếp theo

ngu người đê anh em

AD BDT Cosi ta co : 

a) y=\(\frac{x}{2}+\frac{18}{x}>=2\sqrt{\frac{x18}{2x}}=2\sqrt{9}=6\)

vay GTNN cua y la 6 khi x>0

b) y=\(\frac{x}{2}+\frac{2}{x-1}>=2\sqrt{\frac{x2}{2x-2}}=2\cdot-2=-4\)

vay GTNN cua y la -4 khi x>1

cac cau con lai cung ad tuong tu nha cac p 

p nao thay dung thi k mk nka mk se k lai cko

Mình áp dụng luôn Cô - si cho các số ta được

a) \(\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}\cdot\frac{18}{x}}=2.\sqrt{9}=2.3=6\)

b) \(y=\frac{x}{2}+\frac{2}{x-1}=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{x-1}{2}\cdot\frac{2}{x-1}}+\frac{1}{2}=2+\frac{1}{2}=\frac{5}{2}\)

c) \(\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2}\cdot\frac{1}{x+1}}-\frac{3}{2}=2\sqrt{\frac{3}{2}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)

h) \(x^2+\frac{2}{x^2}\ge2\sqrt{x^2\cdot\frac{2}{x^2}}=2\sqrt{2}\)

g) \(\frac{x^2+4x+4}{x}=\frac{\left(x+2\right)^2}{x}\ge0\)

\(y=\frac{x-1}{2}+\frac{2}{x-1}+\frac{1}{2}\ge2\sqrt{\frac{2\left(x-1\right)}{2\left(x-1\right)}}+\frac{1}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi \(\frac{x-1}{2}=\frac{2}{x-1}\Rightarrow x=3\)

\(y=\frac{5\left(3x-1\right)}{9}+\frac{5}{3x-1}+\frac{5}{9}\ge2\sqrt{\frac{25\left(3x-1\right)}{9\left(3x-1\right)}}+\frac{5}{9}=\frac{35}{9}\)

Dấu "=" xảy ra khi \(x=\frac{4}{3}\)

\(y=-2+\frac{2}{1-x}+\frac{3}{x}\ge-2+\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{1-x+x}=3+2\sqrt{6}\)

Dấu "=" xảy ra khi \(\frac{1-x}{\sqrt{2}}=\frac{x}{\sqrt{3}}\Rightarrow x=3-\sqrt{6}\)

\(y=x+\frac{9}{x}+2020\ge2\sqrt{\frac{9x}{x}}+2020=2026\)

Dấu "=" xảy ra khi \(x=3\)

Áp dụng cô si

\(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\\\frac{1}{c}+\frac{1}{b}\ge2\sqrt{\frac{1}{cb}}\\\frac{1}{a}+\frac{1}{c}\ge2\sqrt{\frac{1}{ac}}\end{cases}}\)\(\Rightarrow\frac{1}{c}+\frac{1}{b}+\frac{1}{a}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}}\)

\("="\Leftrightarrow a=b=c=0\)

\(\hept{\begin{cases}\sqrt{x}\le\frac{x+1}{2}\\\sqrt{y-1}\le\frac{y-1+1}{2}\\\sqrt{z-2}\le\frac{z-2+1}{2}\end{cases}}\)\(\Rightarrow\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{x+1+y-1+1+z-2+1}{2}\)

\(\Leftrightarrow\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\le\frac{x+y+z}{2}\)

\("="\Leftrightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)

Sửa ĐK của c) : a, b, c > 0

Áp dụng bất đẳng thức Cauchy ta có :

\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}=\frac{2}{\sqrt{ab}}\)

\(\frac{1}{b}+\frac{1}{c}\ge2\sqrt{\frac{1}{bc}}=\frac{2}{\sqrt{bc}}\)

\(\frac{1}{c}+\frac{1}{a}\ge2\sqrt{\frac{1}{ca}}=\frac{2}{\sqrt{ca}}\)

Cộng các vế tương ứng

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\ge\frac{2}{\sqrt{ab}}+\frac{2}{\sqrt{bc}}+\frac{2}{\sqrt{ca}}\)

=> \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

=> đpcm

Đẳng thức xảy ra khi a = b = c

c) Cách khác: Áp dụng bổ đề: \(x^2+y^2+z^2\ge xy+yz+zx\forall x,y,z>0\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2+\left(\frac{1}{\sqrt{c}}\right)^2\ge\frac{1}{\sqrt{a}}.\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}.\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}.\frac{1}{\sqrt{a}}\)

\(=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

Dấu "=" xảy ra khi \(a=b=c>0\)

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

Nhân cả 2 vế với a+b+c 

Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)

\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0

dễ rồi nhé

b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được: 

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)

=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)

=>P_max=3/4 <=> x=y=z=1/3

c) Áp dụng bđt Cauchy Schwarz dạng Engel ta được:

\(P=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+2ab+b^2+2ac+c^2+2ab}=\frac{9}{\left(a+b+c\right)^2}\)

<=>\(P\ge\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1^2}=9\)

Vậy P_min=9 <=> a=b=c=1/3