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 \(a=b=c\)

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

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 \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)

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

Lời giải:

Ta thấy:

\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)

\(=\frac{a(a+b+c)}{a+b}+\frac{b(a+b+c)}{b+c}+\frac{c(a+b+c)}{c+a}\)

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

\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)

Có:

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$

$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$

Do đó:

$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)

Áp dụng BĐT AM-GM:

\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)

\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=\frac{1}{\sqrt{3}}$

Quay lại diễn đàn trong thinh lặng:))

Chứng minh: $$\left( a+{\frac {ab}{b+c}}+b+{\frac {bc}{c+a}}+c+{\frac {ca}{a+b}}
\right) ^{2}-{\frac {27\,ab}{4}}-{\frac {27\,ca}{4}} \geqq {\frac {27\,bc}{
4}}$$

Sau khi quy đồng, cần chứng minh$:$

$$\frac{1}{2} \sum\limits_{cyc} \left( 5\,{a}^{4}{b}^{2}+8\,{a}^{3}{b}^{3}+7\,{a}^{2}{b}^{4}+98\,{a}^
{2}{b}^{3}c+99\,{a}^{2}{b}^{2}{c}^{2}+124\,{a}^{2}b{c}^{3}+34\,a{b}^{4
}c+130\,a{b}^{3}{c}^{2}+26\,{b}^{4}{c}^{2}+44\,{b}^{3}{c}^{3}+{c}^{6}
\right) \left( a-b \right) ^{2} \geqq 0$$

Xét: \(1+c^2=ab+bc+ca+c^2=\left(a+c\right)\left(b+c\right)\)

Tương tự CM được:

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

Mặt khác ta tách: \(\hept{\begin{cases}a-b=\left(a+c\right)-\left(b+c\right)\\b-c=\left(a+b\right)-\left(c+a\right)\\c-a=\left(c+b\right)-\left(a+b\right)\end{cases}}\)

Thay vào ta được:

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

\(=\frac{1}{b+c}-\frac{1}{c+a}+\frac{1}{c+a}-\frac{1}{a+b}+\frac{1}{a+b}-\frac{1}{b+c}\)

\(=0\)

=> đpcm

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

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

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

Áp dụng bđt Cô-si :

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

Chứng minh tương tự với các phân thức còn lại, cộng theo vế ta có :

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

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

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