Với \(a=b=c=0\Leftrightarrow S=abc=0\)

Với \(a,b,c\ne0\)

Ta có \(\dfrac{a}{1+ab}=\dfrac{b}{1+bc}=\dfrac{c}{1+ac}\Leftrightarrow\dfrac{1+ab}{a}=\dfrac{1+bc}{b}=\dfrac{1+ac}{c}\)

\(\Leftrightarrow\dfrac{1}{a}+b=\dfrac{1}{b}+c=\dfrac{1}{c}+a\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{1}{a}-\dfrac{1}{c}=\dfrac{c-a}{ac}\\b-c=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{ab}\\c-a=\dfrac{1}{c}-\dfrac{1}{b}=\dfrac{b-c}{bc}\end{matrix}\right.\)

Nhân vế theo vế ta đc \(\left(a-b\right)\left(b-c\right)\left(c-a\right)=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{ab\cdot bc\cdot ca}\)

\(\Leftrightarrow\left(abc\right)^2=1\Leftrightarrow\left[{}\begin{matrix}abc=1\\abc=-1\end{matrix}\right.\)

Đặt: \(M=\frac{1}{a+bc}+\frac{1}{b+ca}+\frac{1}{c+ab}=\Sigma_{cyc}\frac{a}{a^2+ab+bc+ca}\)

\(\Rightarrow M.\left(a+b+c\right)=3-\Sigma_{cyc}\frac{bc}{a^2+ab+bc+ca}\)

Đến đây t cần chứng minh:

 \(\frac{bc}{a^2+ab+bc+ca}+\frac{ca}{b^2+ab+bc+ca}+\frac{ab}{c^2+ab+bc+ca}\ge\frac{3}{4}\) (*)

Từ điều kiện ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\left(x,y,z>0\right)\)

\(\Rightarrow x+y+z=1\)

(*) \(\Leftrightarrow\frac{x^2}{\left(x+y\right)\left(z+x\right)}+\frac{y^2}{\left(x+y\right)\left(y+z\right)}+\frac{z^2}{\left(y+z\right)\left(z+x\right)}\ge\frac{3}{4}\)

Theo Cô-si: \(\frac{x^2}{\left(x+y\right)\left(z+x\right)}+\frac{9}{16}\left(x+y\right)\left(z+x\right)\ge\frac{3}{2}x\)

Nhứng phần kia tương tự

\(\Rightarrow\Sigma_{cyc}\frac{x^2}{\left(x+y\right)\left(z+x\right)}\ge\frac{3}{2}\left(x+y+z\right)-\frac{9}{16}\left[\left(x+y+z\right)^2+\left(xy+yz+zx\right)\right]\ge\frac{3}{4}\)

Lần trước làm không đúng hy vọng bây giờ gỡ lại được

nub

Bạn suy ra dòng 8 mk chưa hiểu, giải kĩ cho mk đc ko

À hiểu r nha bạn,

Bài làm thật xuất sắc!

1. Đề thiếu

2. BĐT cần chứng minh tương đương:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Ta có:

\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)

3.

Ta có:

\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)

\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)

Lại có:

\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)

4.

Ta có:

\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)

\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

5.

Ta có:

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)

\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)

Câu 1:

\(VT=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(VT=1-\dfrac{1}{n}< 1\) (đpcm)

1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)

\(=ac+bc+c^2+ab\)

\(=a\left(b+c\right)+c\left(b+c\right)\)

\(=\left(b+c\right)\left(a+b\right)\)

CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)

\(b+ca=\left(b+c\right)\left(a+b\right)\)

Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)

Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)

CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)

\(\Rightarrow P\le\frac{1}{2}.3\)

\(\Rightarrow P\le\frac{3}{2}\)

Dấu"="xảy ra \(\Leftrightarrow a=b=c\)

Vậy /...

\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)

\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)

Tương tự rồi cộng lại:

\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)

\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)

Dấu "=" xảy ra tại \(a=b=c=1\)

Bài 1 : 

\(P=\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}\)

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

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

\(P=\sqrt{\frac{ab}{ac+bc+c^2+ab}}+\sqrt{\frac{bc}{a^2+ab+ac+bc}}\)

\(+\sqrt{\frac{ca}{ab+b^2+bc+ca}}\)

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

Áp dụng bất đẳng thức Cauchy cho 2 bô só thực không âm

\(\Rightarrow\hept{\begin{cases}\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{2}\\\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{a}{a+b}+\frac{c}{b+c}}{2}\end{cases}}\)

\(\Rightarrow VT\)

\(\le\frac{\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{b}{a+b}+\frac{a}{a+b}\right)}{2}\)

\(\Rightarrow VT\le\frac{\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)

\(\Rightarrow P\le\frac{3}{2}\)

Vậy \(P_{max}=\frac{3}{2}\)

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)

Chúc bạn học tốt !!!

\(HUY=\frac{abc}{abc+a+ab}+\frac{1}{1+b+bc}+\frac{b}{b+bc+abc}=\frac{bc}{bc+1+b}+\frac{1}{1+b+bc}+\frac{b}{b+bc+1}=1\)