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

\(=a^2b^2c^2+a^2b^2+a^2c^2+a^2+b^2c^2+b^2+c^2+1=\)

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

Ta có

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

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

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1-2abc\left(a+b+c\right)\) (2)

Ta có

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

\(=a^2+b^2+c^2+2\)

\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\) (3)

Thay (2) và (3) vào (1)

\(Q=a^2b^2c^2+1-2abc\left(a+b+c\right)+\left(a+b+c\right)^2-2+1=\)

\(=\left(abc\right)^2-2abc\left(a+b+c\right)+\left(a+b+c\right)^2=\)

\(=\left[abc-\left(a+b+c\right)\right]^2\)

\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)

\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)

Theo bđt Cauchy - Schwart ta có:

\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)

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

Đặt \(ab+bc+ca=x;abc=y\).

Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)

\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )

Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1

sai rồi nhé bạn 

làm sao mà \(x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\)lại luôn đúng

Thay ab+bc+ac = 1 vào Q

Thay ab+bc+ac = 1 và Q ta được :

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

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

\(=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) là bình phương  của một số hữu tỉ (đpcm)

Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\) 

Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

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

Ta có:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)

\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{6a-b-c-2}{8}\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\frac{6b-c-a-2}{8}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6c-a-b-2}{8}\end{cases}}\)

Cộng vế theo vế ta được

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6a-b-c-2}{8}+\frac{6b-c-a-2}{8}+\frac{6c-a-b-2}{8}\)

\(=\frac{a+b+c}{2}-\frac{3}{4}\ge\frac{3}{2}.\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)

Mai mình làm cho

ukm mơn alibaba nguyễn nhìu

Ta chứng minh BĐT sau cho các số dương:

\(x^5+y^5\ge xy\left(x^3+y^3\right)\)

\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)

Áp dụng:

\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)

Tương tự và cộng lại:

\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)

\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)

\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)

Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).

Vì ab+bc+ca=1

\(\Rightarrow a^2+1\)

\(=a^2+ab+bc+ca\)

\(=\left(a^2+ab\right)+\left(ac+bc\right)\)

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

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

Tương tự ta được \(\begin{cases}b^2+1=\left(b+a\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(c+b\right)\end{cases}\)

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

\(=\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)

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

Mặt khác a;b;c là số hữa tỉ

\(\Rightarrow\begin{cases}a+b\\b+c\\c+a\end{cases}\) là số hữu tỉ

\(\Rightarrow\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) là số hữu tỉ

=> đpcm