1, Ta có \(abc=a+b+c+2\ge4\sqrt[4]{abc.2}\)

<=>\(abc\ge8\)

BĐT <=> \(ab+bc+ac\ge2\left(abc-2\right)\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2-\frac{2}{abc}\)

Áp dụng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\)

Khi đó cần CM \(\frac{3}{\sqrt[3]{abc}}\ge2-\frac{2}{abc}\)

Đặt \(\frac{1}{\sqrt[3]{abc}}=x\)=> \(0< x\le2\)

BĐT<=> \(\frac{3}{x}\ge2-\frac{2}{x^3}\)

<=>\(\frac{2}{x^3}+\frac{3}{x}-2\ge0\)

<=> \(2+3x^2-2x^3\ge0\)

<=> \(\left(2-x\right)\left(2x^2+x+1\right)\ge0\)(luôn đúng với \(0< x\le2\))

=> BĐT được CM

Dấu bằng xảy ra khi a=b=c=2

2. BĐT <=> \(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}}\le\frac{3}{2}\)

Đặt \(a=\frac{y+z}{x};b=\frac{x+z}{y}\left(x.y,z>0\right)\)

=> \(c=\frac{a+b+2}{ab-1}=\frac{\frac{y+z}{x}+\frac{x+z}{y}+2}{\frac{\left(y+z\right)\left(x+z\right)}{xy}-1}=\frac{x^2+y^2+z\left(x+y\right)+2xy}{z\left(x+y+z\right)}=\frac{\left(x+y\right)^2+z\left(x+y\right)}{z\left(x+y+z\right)}=\frac{x+y}{z}\)

Khi đó BĐT <=> \(\frac{1}{\sqrt{\frac{\left(y+z\right)\left(x+z\right)}{xy}}}+\frac{1}{\sqrt{\frac{\left(x+z\right)\left(x+y\right)}{yz}}}+\frac{1}{\sqrt{\frac{\left(x+y\right)\left(y+z\right)}{zx}}}\le\frac{3}{2}\)

<=> \(\sqrt{\frac{xy}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{yz}{\left(x+z\right)\left(x+y\right)}}+\sqrt{\frac{xz}{\left(y+z\right)\left(x+y\right)}}\le\frac{3}{2}\)

Áp dụng cosi ta có 

\(\sqrt{\frac{xy}{\left(y+z\right)\left(x+z\right)}}\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)\)

Tương tự=> \(VT\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{z}{x+z}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{y}{x+y}\right)=\frac{3}{2}\)(ĐPCM)

Dấu bằng xảy ra khi \(x=y=z\)=> \(a=b=c=2\)

Xin lỗi ở bài 1 mình làm nhầm .

Làm lại bài1

Tương tự bài 2

Đặt \(a=\frac{y+z}{x},b=\frac{x+z}{y},c=\frac{x+y}{z}\left(x,y,z>0\right)\)

Khi đó BĐT

<=> \(\frac{\left(y+z\right)\left(x+z\right)}{xy}+\frac{\left(x+z\right)\left(x+y\right)}{yz}+\frac{\left(y+z\right)\left(x+y\right)}{xz}\ge2\left(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}\right)\)

<=> \(z\left(y+z\right)\left(x+z\right)+x\left(x+y\right)\left(x+z\right)+y\left(y+z\right)\left(y+x\right)\ge2\left[yz\left(y+z\right)+xz\left(x+z\right)+xy\left(x+y\right)\right]\)

<=>\(x^3+y^3+z^3+3xyz\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)( BĐT Schur ) (luôn đúng)

=> BĐT được CM

Dấu bằng xảy ra khi a=b=c=2

\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)

Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)

Lời giải:
ĐKĐB như thế này sinh ra 1 đẳng thức rất đẹp.

$a+b+c+2=abc\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1$

Khi đó, áp dụng BĐT Bunhiacopxky:

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

\(\Leftrightarrow \left(\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}\right)\geq \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)^2\)

\(\Leftrightarrow 3+\sum \frac{1}{a}\geq \sum \frac{1}{a}+2\sum \frac{1}{\sqrt{ab}}\Leftrightarrow \sum \frac{1}{\sqrt{ab}}\leq \frac{3}{2}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

Đề: \(\frac{1}{\sqrt{a^4-a^3+ab+2}}+\frac{1}{\sqrt{b^4-b^3+bc+2}}+\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\sqrt{3}\) ???

*Ta chứng minh : \(x^4-x^3+2\ge x+1\forall x>0\)

\(\Leftrightarrow x^4-x^3-x+1\ge0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)\ge0\) ( đúng )

Do đó: \(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\) \(\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}=\sqrt{3}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)

ko biết

Ta có: \(\frac{1}{\sqrt{1+a^2}}=\sqrt{\frac{abc}{abc+a^2\left(a+b+c\right)}}=\sqrt{\frac{bc}{ac+a^2+ab+ac}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\)

Áp dụng bđt Cô-si được

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

Thiết lập các bđt còn lại cho 2 số hạng còn lại rồi cộng vào được đpcm

bạn có đang on không chat vs mình đi

Trước khi đọc lời giải hãy thăm nhà em trước nhé ! See method from solution! Cảm ơn mn!

Ok, giờ chú ý:

\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\)

\(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{ab.ca+abc+ab}\)

\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}=1\) với abc = 1.

Như vậy: \(VT=\sqrt{\left(\Sigma\frac{1}{\sqrt{ab+a+2}}\right)^2}\le\sqrt{3\left(\Sigma\frac{1}{\frac{\left(ab+a+1\right)}{3}+\frac{\left(ab+a+1\right)}{3}+\frac{\left(ab+a+1\right)}{3}+1}\right)}\)

\(\le\sqrt{\frac{3}{16}\left[\Sigma\left(\frac{9}{ab+a+1}+1\right)\right]}=\frac{3}{2}\)

Đẳng thức xảy ra khi a = b = c = 1

\(\frac{a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)

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

\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)

\(\le\frac{1}{2}.\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{c}{c+b}\right)=\frac{3}{2}\)