b) với mọi a,b,c ϵ R và x,y,z ≥ 0 có :
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(1\right)\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Thật vậy với a,b∈ R và x,y ≥ 0 ta có:
\(\frac{a^2}{x}=\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(2\right)\)
⇔\(\frac{a^2y}{xy}+\frac{b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\)
⇔\(\frac{a^2y+b^2x}{xy}.\left(x+y\right)xy\ge\frac{\left(a+b\right)^2}{x+y}.\left(x+y\right)xy\)
⇔\(\left(a^2y+b^2x\right)\left(x+y\right)\ge\left(a+b\right)^2xy\)
⇔\(a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+2abxy+b^2xy\)
⇔\(b^2x^2+a^2y^2-2abxy\ge0\)
⇔\(\left(bx-ay\right)^2\ge0\)(luôn đúng )
Áp dụng BĐT (2) có:
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}=\frac{\left(a+b+c\right)^2}{x+y+z}\)
Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)} \)
= \(\frac{1}{a^2}.\frac{1}{ab+ac}+\frac{1}{b^2}.\frac{1}{bc+ac}+\frac{1}{c^2}.\frac{1}{ac+bc}\)
=\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\)
Áp dụng BĐT (1) ta có:
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}++\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}\)
Mà abc=1⇒\(\left\{{}\begin{matrix}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ac=\frac{1}{b}\end{matrix}\right.\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)
\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=3\sqrt[3]{\frac{1}{1}}=3\)( BĐT cosi )
⇒\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)
⇒\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}.3=\frac{3}{2}\)
Vậy \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Chúc bạn học tốt !!!

1.

\(10x=|x+\dfrac{1}{10}|+|x+\dfrac{2}{10}|+...+|x+\dfrac{9}{10}| \ge 0\)

\(\Rightarrow x\ge0\)

\(pt\Leftrightarrow x+\frac{1}{10}+x+\frac{2}{10}+...+x+\frac{9}{10}=10x\)

\(\Leftrightarrow x=\frac{1}{10}+\frac{2}{10}+...+\frac{9}{10}=\frac{9}{2}\)

\(\Rightarrow x=\frac{9}{2}\)

4.

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a}{b+3c}=\frac{b}{c+3a}=\frac{c}{a+3b}=\frac{a+b+c}{4\left(a+b+c\right)}=\frac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}4a=b+3c\left(1\right)\\4b=c+3a\left(2\right)\\4c=a+3b\left(3\right)\end{matrix}\right.\)

Từ \(\left(1\right);\left(2\right)\Rightarrow4a=b+3\left(4b-3a\right)\)

\(\Rightarrow12a=12b\Rightarrow a=b\left(4\right)\)

Từ \(\left(1\right);\left(3\right)\Rightarrow4c=a+3\left(4a-3c\right)\)

\(\Rightarrow12a=12c\Rightarrow a=c\left(5\right)\)

Từ \(\left(4\right);\left(5\right)\Rightarrow a=b=c\left(đpcm\right)\)

a, \(M=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^n}\)

\(\Rightarrow4M=1+\frac{1}{4}+...+\frac{1}{4^{n-1}}\)

\(\Rightarrow3M=1-\frac{1}{4^n}< 1\Rightarrow M< \frac{1}{3}\left(đpcm\right)\)

b, Lập luận tương tự câu a

\(M=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^n}\)

\(\Rightarrow4M=1+\frac{1}{4}+...+\frac{1}{4^{n-1}}\)

\(\Rightarrow3M=1-\frac{1}{4^n}< 1+\frac{1}{3}=\frac{4}{3}\Rightarrow M< \frac{4}{9}\left(đpcm\right)\)

bài 1 a, hình như có thêm đk là a+b+c=3

Bài 4 nha

Áp dụng BĐT cô si ta có

\(\frac{1}{x^2}+x+x\ge3\sqrt[3]{\frac{1}{x^2}.x.x}=3.\)

Tương tự với y . \(A\ge6\)dấu = xảy ra khi x=y=1

câu 1 mk bị lộn nhưng đáng ra  ca^2 thành c^2a  mới đúng

Câu 1: Đặt   \(S=\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}=\frac{x}{\sqrt{\left(1-x\right)\left(x+1\right)}}+\frac{y}{\sqrt{\left(1-y\right)\left(y+1\right)}}\)

\(\frac{S}{\sqrt{3}}=\frac{x}{\sqrt{\left(3-3x\right)\left(x+1\right)}}+\frac{y}{\sqrt{\left(3-3y\right)\left(y+1\right)}}\)

Áp dụng BĐT AM-GM: \(\sqrt{\left(3-3x\right)\left(x+1\right)}\le\frac{3-3x+x+1}{2}=\frac{4-2x}{2}=2-x\)

\(\Rightarrow\frac{x}{\sqrt{\left(3-3x\right)\left(x+1\right)}}\ge\frac{x}{2-x}\)

Tương tự: \(\frac{y}{\sqrt{\left(3-3y\right)\left(y+1\right)}}\ge\frac{y}{2-y}\)

Từ đó: \(\frac{S}{\sqrt{3}}\ge\frac{x}{2-x}+\frac{y}{2-y}=\frac{x^2}{2x-x^2}+\frac{y^2}{2y-y^2}\)

Áp dụng BĐT Schwarz: \(\frac{S}{\sqrt{3}}\ge\frac{x^2}{2x-x^2}+\frac{y^2}{2y-y^2}\ge\frac{\left(x+y\right)^2}{2\left(x+y\right)-\left(x^2+y^2\right)}=\frac{1}{2-\left(x^2+y^2\right)}\)

Áp dụng BĐT \(\frac{x^2+y^2}{2}\ge\frac{\left(x+y\right)^2}{4}\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{S}{\sqrt{3}}\ge\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\Leftrightarrow S\ge\frac{2\sqrt{3}}{3}=\frac{2}{\sqrt{3}}\)(ĐPCM).

Dấu bằng có <=> \(x=y=\frac{1}{2}\).

Câu 4: Sửa đề CMR: \(abcd\le\frac{1}{81}\)

 Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=3\)

\(\Leftrightarrow\frac{1}{1+a}=\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)

\(\Leftrightarrow\frac{1}{1+a}=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)(AM-GM)

Tương tự: 

\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)\(;\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)

\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Nhân 4 BĐT trên theo vế thì có: 

\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^3}}\)

\(=81.\frac{abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)

\(\Rightarrow81.abcd\le1\Leftrightarrow abcd\le\frac{1}{81}\)(ĐPCM)

Dấu "=" có <=> \(a=b=c=d=\frac{1}{3}\).

Úi mơn bạn nhiều

<=> \(\frac{m^2y+n^2x}{xy}>=\left(\frac{m^2+2mn+n^2}{x+y}\right)\)

<=> \(\left(m^2y+n^2x\right).\left(x+y\right)>=\left(m^2+2mn+n^2\right).xy\)(vì x,y,m^2,n^2 >= 0)

<=> m²xy + n²xy + m²y² + n²x² >= m²xy + n²xy + 2mnxy

<=> n²x² + m²y² >= 2mnxy (luôn đúng) (bất đẳng thức cosi).

Vậy ....

Bài 1 :

Từ \(\left(x+1\right)\left(y+1\right)=4xy\)

\(\Rightarrow\frac{x+1}{x}.\frac{y+1}{y}=4\Leftrightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=4\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y}\), ta có :

\(\left(1+a\right)\left(1+b\right)=4\Leftrightarrow3=a+b+ab\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+2\sqrt{ab}+ab\ge2\sqrt{ab}+ab\)

Từ đó \(ab\le1\)

Áp dụng AM - GM cho 2 số thực dương ta có :
\(\frac{1}{\sqrt{3x^2+1}}=\frac{\frac{1}{x}}{\sqrt{3+\frac{1}{x^2}}}=\frac{a}{\sqrt{a+b+ab+a^2}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}\right)\)

Tương tự ta có :

\(\frac{1}{\sqrt{3y^2+1}}\le\frac{1}{2}.\left(\frac{a}{a+b}+\frac{b}{b+1}\right)\)

Bài 1 :

Vì \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\) nên \(b=\frac{2ac}{a+c}\)

Do đó : \(\frac{a+b}{2a-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}=\frac{c^2+3ac}{2a^2}=\frac{a+3c}{2a}\)

Và : \(\frac{c+b}{2c-b}=\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}=\frac{c^2+3ac}{2c^2}=\frac{c+3a}{2c}\)

Suy ra \(P=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+3c}{2a}+\frac{c+3a}{2c}=\frac{ac+3c^2+ac+3a^2}{2ac}\)

\(=\frac{3\left(a^2+c^2\right)+2ac}{2ac}\ge\frac{3.2ac+2ac}{2ac}=\frac{8ac}{2ac}=4\)

Vậy \(P\ge4\) với mọi a,b,c thỏa mãn đề bài. Dấu bằng xảy ra khi a=b=c

Vậy GTNN của P là 4 khi a=b=c

Chúc bạn học tốt !!

Mình nhầm tí nhé bạn dổi bài 1 thành bài 2 nhé ạ sorry bạn nhìu

Áp dụng BĐT Côsi dưới dạng engel, ta có:

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

⇒\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(x+y+z\right).\frac{9}{x+y+z}\) = 9

Dấu "=" xảy ra ⇔ x = y = z

\(P=\frac{1}{x^2+y^2+z^2}+\frac{2009}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}+\frac{2007}{xy+yz+zx}\)

\(P\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{2007}{\frac{1}{3}\left(x+y+z\right)^2}\)

\(P\ge\frac{9}{\left(x+y+z\right)^2}+\frac{6021}{\left(x+y+z\right)^2}=\frac{6030}{\left(x+y+z\right)^2}\ge\frac{6030}{3^2}=670\)

Dấu "=" xảy ra khi \(x=y=z=1\)

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

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

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

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

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

\(=\frac{-\left(x-y\right)\left(x+y\right)-\left(x-y\right)}{x^2y^2+xy\left(x+y\right)+xy+x^2+y^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

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

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

\(=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)