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 !!!

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)

Áp dụng bất đẳng thức Cauchy 

\(\Rightarrow VT\ge3\sqrt[6]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\)

Chứng minh : \(3\sqrt[6]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\ge3\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\)

Áp dụng bất đẳng thức Cauchy 

\(\Rightarrow\left(c+ab\right)\left(a+bc\right)\le\frac{\left(c+a+ab+bc\right)^2}{4}\)

\(=\frac{\left[b\left(a+c\right)+c+a\right]^2}{4}=\frac{\left(b+1\right)^2\left(c+a\right)^2}{4}\)

Thiết lập tương tự và thu lại ta có : 

\(\Rightarrow\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\)

\(\le\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a^2\right)\left(b+1\right)^2\left(a+1\right)^2\left(c+1\right)^2}{64}\)

\(\Rightarrow64\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\)

\(\le\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(c+1\right)^2\left(a+1\right)^2\)

\(\Leftrightarrow8\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\)

\(\le\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(b+1\right)\left(c+1\right)\left(a+1\right)\) 

Cần chứng minh : 

\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le8\)

Áp dụng bất đẳng thức Cauchy 

\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\left(\frac{3+3}{3}\right)^3=8\left(đpcm\right)\)

Chúc bạn học tốt !!!!

\(A=\frac{\frac{1}{2}a^2\left(\sqrt[3]{b}+\sqrt[3]{c}+1\right)\left[\left(\sqrt[3]{b}-\sqrt[3]{c}\right)^2+\left(\sqrt[3]{b}-1\right)^2+\left(\sqrt[3]{c}-1\right)^2\right]}{2\left(a+2\right)\left(a+\sqrt[3]{bc}\right)}\ge0\)

\(\Sigma_{cyc}\frac{a^2}{a+\sqrt[3]{bc}}=\Sigma_{cyc}A+\Sigma_{cyc}\frac{2\left(a-1\right)^2}{3\left(a+2\right)}+\frac{5}{6}\left(a+b+c\right)-1\ge\frac{5}{6}\left(a+b+c\right)-1=\frac{3}{2}\)

Áp dụng bất đẳng thức cộng mẫu số 

\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\)\(\ge\frac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\)\(\ge\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

Chứng minh rằng : \(\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\frac{3}{2}\)

\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)

\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)

\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)

Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm

\(\Rightarrow\hept{\begin{cases}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{cases}}\)

\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)

\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\left(đpcm\right)\)

Vì \(\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\frac{3}{2}\)

Mà \(\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{3}{2}\left(đpcm\right)\)

Chúc bạn học tốt !!!

1,

\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)

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

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

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

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

Áp dụng bất đẳng thức Cô si nhưng tình huống này ta bình phương hai vế trước.

Đặt \(A=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\), khi đó ta được:

\(A^2=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^2\)

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

Ta chú ý cách ghép cặp sau:

\(\frac{x^4}{y^2}=\frac{x^2y}{z}+\frac{x^2y}{x}+z^2\ge4x^2\)

\(\frac{y^4}{z^2}+\frac{y^2z}{x}+\frac{y^2z}{x}+x^2\ge4y^2\)

\(\frac{z^4}{x^2}=\frac{z^2x}{y}+\frac{z^2x}{y}+y^2\ge4z^2\)

Cộng theo vế các bất đẳng thức trên ta được:

\(A^2+\left(x^2+y^2+z^2\right)\ge4\left(x^2+y^2+z^2\right)\Leftrightarrow A^2\ge9\Leftrightarrow A\ge3\)hay:

\(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge3\)

Vậy bất đẳng thức đã được chứng minh, đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)