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

\(\ge\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{9\left(a^2+b^2+c^2\right)}{ab+bc+ca}+2\left(ab+bc+ca\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Leftrightarrow P\ge\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}+18\)

\(\ge2+8+18=28\)

Ta có  \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.1=3\)  \(\Rightarrow a+b+c\ge\sqrt{3}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel

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

Đẳng thức xảy ra  \(\Leftrightarrow\)  \(\hept{\begin{cases}\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\\ab+bc+ca=1\end{cases}}\)  \(\Leftrightarrow\)  \(a=b=c=\frac{\sqrt{3}}{3}\)

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

giải hộ nha gấp

mk thấy khó à nha

Nhớ làm đâu đó rồi mà làm biếng lục vc:(

Đặt \(\left(a+b+c;ab+bc+ca;abc\right)=\left(3u;3v^2;w^3\right)\). Ta đi chứng minh \(P\ge28\)

\(\Leftrightarrow\frac{v^2}{3u^2-2v^2}+\frac{27u^3}{w^3}\ge28\). Chú ý rằng: \(w^3\le uv^2\). Do đó ta chỉ cần chứng minh:

\(\Leftrightarrow\frac{v^2}{3u^2-2v^2}+\frac{27u^2}{v^2}\ge28\)\(\Leftrightarrow\frac{3\left(u^2-v^2\right)\left(27u^2-19v^2\right)}{v^2\left(3u^2-2v^2\right)}\ge0\)

Hiển nhiên đúng do \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow u^2\ge v^2\)...

P/s: Bài này dùng SOS đi cho lẹ:D

Cách 2:

\(P-28=\frac{\left(a+b+c\right)^2\left[\Sigma_{cyc}a\left(b-c\right)^2\right]}{abc\left(ab+bc+ca\right)}+\frac{\left(\Sigma_{cyc}a^2-\Sigma_{cyc}ab\right)\left(9\Sigma_{cyc}a^2-\Sigma_{cyc}ab\right)}{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}\ge0\)

Vậy \(P\ge28\). Đẳng thức xảy ra khi \(a=b=c\)

Ngoài ra dùng dồn biến cũng ra:D

Chứng minh: \(P=F\left(a;b;c\right)\ge F\left(\frac{a+b}{2};\frac{a+b}{2};c\right)\ge28\)

\(P=\frac{2018}{a^2+b^2+c^2}+\frac{2018}{ab+bc+ac}-\frac{2017}{a^2+b^2+c^2}\)

\(P\ge2018\left(\frac{4}{a^2+b^2+c^2+ab+bc+ac}\right)-\frac{2017}{a^2+b^2+c^2}\)

\(P\ge\frac{2018.8}{\left(a+b+c\right)^2}-\frac{2017}{a^2+b^2+c^2}=\frac{2018.8}{9}-\frac{2017}{a^2+b^2+c^2}\)

Vì \(9=\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2\ge3\)

\(P\ge\frac{2018.8}{9}-\frac{2017}{3}=...\)

P min = ... khi a=b=c = 1