đúng rồi

 chó điên

\(\frac{ab}{a+b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\) 

Làm tương tự với 2 phân thức còn lại rồi cộng vào ra đpcm 

\(\frac{1}{2a-1}+\frac{1}{1}\ge\frac{4}{2a}=\frac{2}{a}\) ; \(\frac{1}{2b-1}+\frac{1}{1}\ge\frac{2}{b}\) ; \(\frac{1}{2c-1}+\frac{1}{1}\ge\frac{2}{c}\)

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

\(\Rightarrow VT\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Sửa đề:

Cho a, b, c > 1(chỗ này là ý tui, dùng Wolfram Alpha sẽ thấy nếu không sửa như vầy thì đẳng thức không xảy ra). CMR:

\(\frac{1}{2a-1}+\frac{1}{2b-1}+\frac{1}{2c-1}+3\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\) (cái này là ý chủ tus đấy nhá!)

\(\Leftrightarrow\frac{2a}{2a-1}+\frac{2b}{2b-1}+\frac{2c}{2c-1}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\) (tách ghép vế trái + làm chặt BĐT do \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b};..\))

\(\Leftrightarrow\frac{2a^2-4a+2}{a\left(2a-1\right)}+\frac{2b^2-4b+2}{b\left(2b-1\right)}+\frac{2c^2-4c+1}{c\left(2c-1\right)}\ge0\) (chuyển vế + quy đồng)

\(\Leftrightarrow\frac{2\left(a-1\right)^2}{a\left(2a-1\right)}+\frac{2\left(b-1\right)^2}{b\left(2b-1\right)}+\frac{2\left(c-1\right)^2}{c\left(2c-1\right)}\ge0\) (đúng)

Đẳng thức xảy ra khi a = b = c = 1

Vậy ta có đpcm.

\(\frac{1}{2a-1}+1\ge\frac{\left(1+1\right)^2}{2a-1+1}=\frac{4}{2a}=\frac{2}{a}\)

HSG Bắc Ninh 2018-2019

Có \(\frac{1}{2a-1}\ge\frac{1}{a^2}\);\(\frac{1}{2b-1}\ge\frac{1}{b^2}\);\(\frac{1}{2c-1}\ge\frac{1}{c^2}\)

\(\Rightarrow\text{Σ}_{cyc}\frac{1}{2a-1}+3\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+3\)

Lại có \(\frac{1}{a^2}+\frac{1}{b^2}\ge^{co-si}\frac{2}{ab}\ge\frac{8}{\left(a+b\right)^2}\)

\(\frac{8}{\left(a+b\right)^2}+2\ge\frac{8}{a+b}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+2\ge\frac{8}{a+b}\)

Tương tự ta có \(\frac{1}{b^2}+\frac{1}{c^2}+2\ge\frac{8}{b+c}\);\(\frac{1}{c^2}+\frac{1}{a^2}+2\ge\frac{8}{c+a}\)

\(\Rightarrow2\left(\text{Σ}_{cyc}\frac{1}{a^2}+3\right)\ge\text{Σ}_{cyc}\frac{8}{a+b}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+3\ge\text{Σ}_{cyc}\frac{4}{a+b}\)

\(\RightarrowĐPCM\left("="\Leftrightarrow a=b=c=1\right)\)

\(GT\Rightarrow\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}=3\)

Ta có: \(\frac{1}{a^4}+\frac{1}{a^4}+\frac{1}{a^4}+\frac{1}{b^4}\ge4\sqrt[4]{\frac{1}{a^{12}b^4}}=\frac{4}{a^3b}\)

Tương tự: \(\frac{3}{b^4}+\frac{1}{c^4}\ge\frac{4}{b^3c}\) ; \(\frac{3}{c^4}+\frac{1}{a^4}\ge\frac{4}{c^3a}\)

\(\Rightarrow\frac{1}{a^3b}+\frac{1}{b^3c}+\frac{1}{c^3a}\le\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}=3\)

\(VT=\frac{1}{a^3b+c^2+c^2+1}+\frac{1}{b^3c+a^2+a^2+1}+\frac{1}{c^3a+b^2+b^2+1}\)

\(VT\le\frac{1}{16}\left(\frac{1}{a^3b}+\frac{2}{c^2}+1+\frac{1}{b^3c}+\frac{2}{a^2}+1+\frac{1}{c^3a}+\frac{2}{b^2}+1\right)\)

\(VT\le\frac{1}{16}\left(\frac{1}{a^3b}+\frac{1}{b^3c}+\frac{1}{c^3a}+2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+3\right)\)

\(VT\le\frac{1}{16}\left(6+2\sqrt{3\left(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right)}\right)=\frac{1}{16}\left(6+6\right)=\frac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)