1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)

\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)

\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)

\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)

\(\frac{a^3}{\left(b+2c\right)^2}+\frac{b+2c}{27}+\frac{b+2c}{27}\ge\frac{a}{3}\)\(\Leftrightarrow\)\(\frac{a^3}{\left(b+2c\right)^2}\ge\frac{1}{3}a-\frac{2}{27}b-\frac{4}{27}c\)

tương tự rồi cộng lại

\(a^2+4\left(b+c\right)^2-bc=4a\left(b+c\right)\)

\(\Rightarrow\left[a-2\left(b+c\right)\right]^2=bc\)

Do \(\left(b,c\right)=1\) và \(b.c\) là 1 số chính phương

\(\Rightarrow b,c\) đều là các số chính phương

TA CÓ:

\(a^4b^2+b^4c^2\ge2a^2b^3c,b^4c^2+c^4a^2\ge2b^2c^3a,c^4a^2+a^4b^2\ge2c^2a^3b\)

\(\Rightarrow a^4b^2+b^4c^2+c^4a^2+\frac{5}{9}\ge a^2b^3c+b^2c^3a+c^2a^3b+\frac{5}{9}\)

ĐẶT \(ab=x,bc=y,ca=z\Rightarrow x+y+z=1\)

\(\Rightarrow a^2b^3c+b^2c^3a+c^2a^3b+\frac{5}{9}=x^2y+y^2z+z^2x+\frac{5}{9}\)

TA CẦN C/M:

\(x^2y+y^2z+z^2x+\frac{5}{9}\ge2\left(xy+yz+zx\right)\)        \(\left(=2abc\left(a+b+c\right)\right)\)

ÁP DỤNG BĐT BUNHIA TA CÓ:

\(\left(x^2y+y^2z+z^2x\right)\left(x+y+z\right)\ge\left(xy+yz+zx\right)^2\) DO:\(\left(x+y+z=1\right)\)

VẬY CẦN C/M:

\(\left(xy+yz+zx\right)^2+\frac{5}{9}\ge2\left(xy+yz+zx\right)\)

XÉT HIỆU:

\(\left(xy+yz+zx\right)^2-2\left(xy+yz+zx\right)+1-\frac{4}{9}=\left(xy+yz+zx-1\right)^2-\frac{2^2}{3^2}\)

\(=\left(xy+yz+zx-\frac{1}{3}\right)\left(xy+yz+zx-\frac{5}{3}\right)\)

VÌ:

\(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=\frac{1}{3}\Leftrightarrow xy+yz+zx-\frac{1}{3}\le0\)

\(\Rightarrow\left(xy+yz+zx-\frac{1}{3}\right)\left(xy+yz+zx-\frac{5}{3}\right)\ge0\)

\(\Rightarrow DPCM\)

Bài này mình có hỏi trên mạng ấy bạn bài này nhiều cách lắm tại mình thấy cách này dễ hiểu nên gửi cho b

Giả sử \(c=min\left\{a,b,c\right\}\)

Ta viết BĐT lại thành:\(\frac{5}{9}\left(ab+bc+ca\right)^3+a^4b^2+b^4c^2+c^4a^2\ge2abc\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(VT-VP=(a-b)^2(a^2c^2+\frac{17}{9}abc^2+b^2c^2+\frac{5}{9}ac^3+\frac{5}{9}bc^3)+(a-c)(b-c)(a^3b+\frac{5}{9}a^2b^2+a^3c+\frac{11}{9}a^2bc+\frac{2}{9}ab^2c+a^2c^2)\ge0\)

Áp dụng BĐT BCS dạng phân thức ta được:

\(\frac{1}{4a^2+b^2+c^2}=\frac{1}{9}.\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(c^2+a^2\right)}\le\frac{1}{9}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+a^2}\right)\)

\(\frac{1}{a^2+4b^2+c^2}=\frac{1}{9}.\frac{\left(a+b+c\right)^2}{2b^2+\left(a^2+b^2\right)+\left(c^2+b^2\right)}\le\frac{1}{9}\left(\frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{c^2+b^2}\right)\)

\(\frac{1}{a^2+b^2+4c^2}=\frac{1}{9}.\frac{\left(a+b+c\right)^2}{2c^2+\left(a^2+c^2\right)+\left(c^2+b^2\right)}\le\frac{1}{9}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{c^2+b^2}+\frac{c^2}{2c^2}\right)\)

Cộng các BĐT trên theo vế ta được:

\(\frac{1}{4a^2+b^2+c^2}+\frac{1}{a^2+4b^2+c^2}+\frac{1}{a^2+b^2+4c^2}\le\frac{1}{2}\)

Dấu $"="$ xảy ra \(\Leftrightarrow a=b=c\)

Băng :v

bai nay co gi kho dau nhi <(")