\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)

\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)

Áp dụng BĐT Cosi ta có:

\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)

Từ (1)(2)(3) ta có:

\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)

Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)

Dấu "=" xảy ra <=> a=b=c=1

CHÚC BAN HỌC GIỎI

đây\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

vào thống kê để xem hình ảnh

moi nguoi oi hom truoc minh hoc tap hop cac so TN do thi co cua minh day nhu sau 

vd: A={xeN/3<x<9}

thi minh liet ke ra la A=4,5,6,7,8 nhung sua bai lai ko dung 

co sua nhu vay A=3,4,5,6,7,8

ko biet hay sai mong ae giup minh

Áp dụng BĐT Cô-si \(ab\le\frac{\left(a+b\right)}{4}^2\)

=> \(\left(2a+b\right)\left(2c+b\right)\le\frac{4\left(a+b+c\right)^2}{4}=\left(a+b+c\right)^2\)

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

Mấy cái kia làm tương tự cậu nhé 

Dấu "=" xảy ra khi và chỉ khi a=b=c=1

24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13

Tìm GTNN a: $F= 14(a^2+b^2+c^2) + \dfrac{ab+bc+ca}{a^2b+b^2c+c^2a}$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam

Ta có:

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

\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2b+b^2c+c^2a\right)\le\frac{\left(a^2+b^2+c^2\right)^3}{3}\le\left(a^2+b^2+c^2\right)^4\)

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

Ta lại có:

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

Làm tiếp.

1. Ta có: \(ab+bc+ca=3abc\)

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

Đặt \(\hept{\begin{cases}\frac{1}{a}=m\\\frac{1}{b}=n\\\frac{1}{c}=p\end{cases}}\) khi đó \(\hept{\begin{cases}m+n+p=3\\M=2\left(m^2+n^2+p^2\right)+mnp\end{cases}}\)

Áp dụng Cauchy ta được:

\(\left(m+n-p\right)\left(m-n+p\right)\le\left(\frac{m+n-p+m-n+p}{2}\right)^2=m^2\)

\(\left(n+p-m\right)\left(n+m-p\right)\le n^2\)

\(\left(p-n+m\right)\left(p-m+n\right)\le p^2\)

\(\Rightarrow\left(m+n-p\right)\left(n+p-m\right)\left(p+m-n\right)\le mnp\)

\(\Leftrightarrow m^3+n^3+p^3+3mnp\ge m^2n+mn^2+n^2p+np^2+p^2m+pm^2\)

\(\Leftrightarrow\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-pm\right)+6mnp\ge mn\left(m-n\right)+np\left(n-p\right)+pm\left(p-m\right)\)

\(=mn\left(3-p\right)+np\left(3-m\right)+pm\left(3-n\right)\)

\(\Leftrightarrow3\left(m^2+n^2+p^2\right)-3\left(mn+np+pm\right)+6mnp\ge3\left(mn+np+pm\right)-3mnp\)

\(\Leftrightarrow3\left(m^2+n^2+p^2\right)+9mnp\ge6\left(mn+np+pm\right)\)

\(\Leftrightarrow xyz\ge\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)

\(\Rightarrow M\ge2\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)

\(=\frac{5}{3}\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)\)

\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m^2+n^2+p^2+2mn+2np+2pm\right)\)

\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m+n+p\right)^2\)

\(\ge\frac{4}{3}\cdot3+\frac{1}{3}\cdot3^2=4+3=7\)

Dấu "=" xảy ra khi: \(m=n=p=1\Leftrightarrow a=b=c=1\)