a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)

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

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

b thiếu đề

@tth_new, @Nguyễn Việt Lâm, @No choice teen, @Akai Haruma

giúp e vs ạ! Cần gấp

Thanks nhiều

Bunhiacopxki:

\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)

\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)

Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)

\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)

\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)

Nhân phá và rút gọn 2 vế:

\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)

Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

Dấu "=" xảy ra khi \(a=b=c\)

cho đề này:

cho a;b;c là các số thực dương thỏa mãn a²+b²+c²=1.CMR:\(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\le\frac{9}{2}\)

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

<=>\(\frac{bc+a^2}{a+b}-a+\frac{ac+b^2}{b+c}-b+\frac{ab+c^2}{a+c}-c\ge0\)

<=>\(\frac{b\left(c-a\right)}{a+b}+\frac{c\left(a-b\right)}{b+c}+\frac{a\left(b-c\right)}{a+c}\ge0\)

<=>\(\frac{b\left(b+c\right)\left(a+c\right)\left(a-c\right)}{\left(a+b\right)\left(c+c\right)\left(a+c\right)}\)+\(\frac{c\left(a+c\right)\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a\left(a+b\right)\left(b-c\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)

<=>\(\frac{b^2c^2-b^2a^2+bc^3-a^2bc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^3c-ab^2c+c^2a^2-b^2c^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)+\(\frac{a^2b^2-a^2c^2+ab^3-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)

<=>\(\frac{bc^3+a^3c+ab^3-a^2bc-ab^2c-abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)

<=>\(\frac{2bc^3+2a^3c+2ab^3-2a^2bc-2ab^2c-2abc^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)>=0

<=>\(\frac{bc\left(c-a\right)^2+ac\left(a-b\right)^2+ab\left(b-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đung voi moi a,b,c >0)

Dấu ''='' xay ra khi a=b=c

Sao lại thế???

đây là đề bài hay bài làm thế?

lay 3-VT la xong ban ak,day la phuongphap dao dau ma

bạn tham khảo nhé : https://olm.vn/hoi-dap/detail/222370673956.html

Mình nghĩ đề nên cho a,b,c dương nếu không thì từ từ mình suy nghĩ

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

Ta có:\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{\left(a+b\right)\left(a+c\right)}=\frac{\left(a-bc\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{\left(a-bc\right)\left(1-a\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\)

\(=\frac{a-a^2-bc+abc}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}=\frac{a-a^2-bc+abc}{1-a-b-c+ab+bc+ca-abc}=\frac{a-a^2-bc+abc}{ab+bc+ca-abc}\)

\(\Rightarrow P=\frac{a+b+c-a^2-b^2-c^2-ab-bc-ca+3abc}{ab+bc+ca-abc}\)

\(P=\frac{1-\left(a+b+c\right)^2+ab+bc+ca+3abc}{ab+bc+ca-abc}\)

\(P=\frac{ab+bc+ca+3abc}{ab+bc+ca-abc}=1+\frac{4abc}{ab+bc+ca-abc}\)

Cần cm:\(\frac{4abc}{ab+bc+ca-abc}\le\frac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)\ge9abc\)(đúng theo AM-GM)

"="<=>a=b=c=1/3

Ta có : a + bc = a ( a + b + c ) + bc = ( a + c ) ( a + b )

BĐT cần chứng minh tương đương với :

\(\frac{a\left(a+b+c\right)-bc}{\left(a+c\right)\left(a+b\right)}+\frac{b\left(a+b+c\right)-ca}{\left(b+c\right)\left(b+a\right)}+\frac{c\left(a+b+c\right)-ab}{\left(c+a\right)\left(c+b\right)}\le\frac{3}{2}\)

\(\left(a^2+ab+ac-bc\right)\left(b+c\right)+\left(ab+b^2+bc-ac\right)\left(a+c\right)+\left(ac+bc+c^2-ab\right)\left(a+b\right)\le\frac{3}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

khai triển ra , ta được :

\(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+6abc\le\frac{3}{2}\left(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\right)+3abc\)

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

\(\Rightarrow a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\ge6abc\)( nhân với -2 thì đổi dấu )

\(\Rightarrow b\left(a^2-2ac+c^2\right)+a\left(b^2-2bc+c^2\right)+c\left(a^2-2ab+b^2\right)\ge0\)

\(\Rightarrow b\left(a-c\right)^2+a\left(b-c\right)^2+c\left(a-b\right)^2\ge0\)     

vì BĐT cuối luôn đúng nên BĐT lúc đầu đúng

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