1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham

Bất đẳng thức tương đương với

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

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

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{a+b+c}{2\sqrt{3\left(ab+bc+ca\right)}}+1\left(1\right)\)

Áp dụng BĐT Bunhiacopxki ta có:

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

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

Ta chứng minh \(\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{a+b+c}{2\sqrt{3\left(ab+bc+ca\right)}}+1\left(2\right)\)

Đặt \(t=\frac{a+b+c}{\sqrt{3\left(ab+bc+ca\right)}}>0\),từ BĐT \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Ta được \(t^2\ge0\Rightarrow t>1\).BĐT (2) viết lại thành 

\(\frac{3t^2}{2}\ge\frac{t}{2}+1\Leftrightarrow\left(t-1\right)\left(3t+2\right)\ge0\) luôn đúng

=>(2) được chứng minh

Từ (1) và (2) => điều phải chứng minh

Đẳng thức xảy ra khi và chỉ khi a=b=c

áp dụng BĐT bunhiacopxki

Với \(a,b,c\ge0\). Khi đó ta có

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

Chứng minh: \(\left(ab+bc+ca\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a^2+b^2+c^2+abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge a^2+b^2+c^2\)\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{ab+bc+ac}\)

Với \(a,b,c\ge0\) ta có

\(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(b+a\right)\left(c+a\right)}}+\sqrt{\frac{ca}{\left(c+b\right)\left(c+a\right)}}\ge1\)

Áp dụng bất đẳng thức AM-GM ta có:

\(\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}=\Sigma\sqrt{\frac{ab\left(2ab+2bc+2ac\right)^2}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ca\right)^2}}\)

\(\ge\Sigma\sqrt{\frac{ab\left[a\left(b+c\right)+b\left(a+c\right)\right]^2}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ac\right)^2}}\)

\(\ge\Sigma\sqrt{\frac{ab.4a\left(b+c\right)b\left(a+c\right)}{4\left(a+c\right)\left(b+c\right)\left(ab+bc+ca\right)^2}}=\Sigma\frac{ab}{ab+bc+ca}\)

Từ đó ta có \(\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\ge\frac{ab+bc+ca}{ab+bc+ca}=1\)

chứng minh bài toán:

Đặt \(\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ac}}=t\ge1\)

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

Từ đây ta chứng minh \(\sqrt{t^2+2}+\frac{3\sqrt{3}}{t}\ge\frac{7\sqrt{2}}{2}\)

Áp dụng bất đẳng thức bunhiacopxki ta có:

\(\sqrt{t^2+2}+\frac{3\sqrt{3}}{t}=\frac{\sqrt{\left(t^2+2\right)\left(6+2\right)}}{2\sqrt{2}}+\frac{3\sqrt{3}}{t}\ge\frac{t\sqrt{6}+2}{2\sqrt{2}}+\frac{3\sqrt{3}}{t}=\left(\frac{t\sqrt{3}}{2}+\frac{3\sqrt{3}}{t}\right)+\frac{\sqrt{2}}{2}\)

Áp dụng bất đẳng thức Cauchy ta đc:

\(\left(\frac{t\sqrt{3}}{2}+\frac{3\sqrt{3}}{t}\right)+\frac{\sqrt{2}}{2}\ge3\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2}\)

Vậy ta có đpcm

1.

Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)

\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

2.

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng vế với vế:

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)

3.

Từ câu b, thay \(c=1\) ta được:

\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)

4.

\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)

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

Dấu "=" xảy ra khi \(a=b=c\)

5.

\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)

Cộng vế với vế:

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

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

1. bđt được viết lại thành

\(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

Theo bđt AM-GM thì :

\(ab+bc\ge2\sqrt{ab\cdot bc}=2\sqrt{ab^2c}=2b\sqrt{ac}\)

Tương tự : \(bc+ca\ge2c\sqrt{ab}\); \(ab+ca\ge2a\sqrt{bc}\)

Cộng vế với vế

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

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

Dấu "=" xảy ra <=> a=b=c

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

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

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 !!!

Do \(a+b+c=1\)  nên :

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

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

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

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

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

Cộng theo vế :
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\left(đpcm\right)\)

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

Chúc bạn học tốt !!!

Từ giả thiết ta suy ra

\(\dfrac{1}{a}+\dfrac{1}{b}+c=3\)

Đặt \(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};c\right)\Rightarrow x+y+z=3\)

\(VT=\dfrac{1}{\sqrt{xy+x+y}}+\dfrac{1}{\sqrt{yz+y+z}}+\dfrac{1}{\sqrt{xz+x+z}}\)

Ta chứng minh: \(\left(x+1+y\right)^2\ge3\left(xy+x+y\right)\)(Luôn đúng)

\(\Rightarrow VT\ge\dfrac{\sqrt{3}}{x+y+1}+\dfrac{\sqrt{3}}{y+z+1}+\dfrac{\sqrt{3}}{z+x+1}\ge\dfrac{9\sqrt{3}}{2\left(x+y+z\right)+3}=\sqrt{3}\)