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\(\Sigma_{sym}a^4b^4\ge\frac{\left(\Sigma_{sym}a^2b^2\right)^2}{3}\ge\frac{\left(\Sigma_{sym}ab\right)^4}{27}\ge\frac{a^2b^2c^2\left(a+b+c\right)^2}{3}=3a^4b^4c^4\)
\(\Sigma\frac{a^5}{bc^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{abc\left(a+b+c\right)}\ge\frac{\left(a^2+b^2+c^2\right)^4}{abc\left(a+b+c\right)^3}\ge\frac{\left(a+b+c\right)^6\left(a^2+b^2+c^2\right)}{27abc\left(a+b+c\right)^3}\)
\(\ge\frac{\left(3\sqrt[3]{abc}\right)^3\left(a^2+b^2+c^2\right)}{27abc}=a^2+b^2+c^2\)
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: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
\(\Leftrightarrow\frac{9}{4a^2+b^2+c^2}+\frac{9}{a^2+4b^2+c^2}+\frac{9}{a^2+b^2+4c^2}\le\frac{9}{2}\)
Thật vậy, ta có:
\(\frac{9}{4a^2+b^2+c^2}=\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\)
Tương tự: \(\frac{9}{a^2+4b^2+c^2}\le\frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{b^2+c^2}\) ; \(\frac{9}{a^2+b^2+4c^2}\le\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{2c^2}\)
Cộng vế với vế:
\(VT\le\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{b^2+c^2}+\frac{a^2}{a^2+c^2}+\frac{c^2}{a^2+c^2}=\frac{3}{2}+3=\frac{9}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Do a ; b ; c > 0 ( GT )
Áp dụng BĐT phụ \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\) , ta có :
\(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow12\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le1\)
Lại có : \(\frac{1}{4a+b+c}=\frac{1}{a+a+a+a+b+c}\le\frac{1}{36}\left(\frac{4}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(1\right)\)
( áp dụng BĐT phụ \(\frac{1}{a1}+\frac{1}{a2}+\frac{1}{a3}+\frac{1}{a4}+\frac{1}{a5}+\frac{1}{a6}\ge\frac{36}{a1+a2+a3+a4+a5+a6}\) )
CMTT , ta có : \(\frac{1}{4b+a+c}\le\frac{1}{36}\left(\frac{4}{b}+\frac{1}{a}+\frac{1}{c}\right);\frac{1}{4c+a+b}\le\frac{1}{36}\left(\frac{4}{c}+\frac{1}{a}+\frac{1}{b}\right)\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow\frac{1}{4a+b+c}+\frac{1}{4b+a+c}+\frac{1}{4c+a+b}\le\frac{1}{36}\left(\frac{6}{a}+\frac{6}{b}+\frac{6}{c}\right)=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{6}.1=\frac{1}{6}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=3\)
a) \(a\ne0;a\ne1\)
\(\Leftrightarrow M=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}\)
\(=\left[\frac{\left(a-1\right)^2}{a^2+a+1}-\frac{1-2a^2+4a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{1}{a-1}\right]\cdot\frac{4a^2}{a\left(a^2+4\right)}\)
\(=\frac{\left(a-1\right)^3-1+2a^2-4a+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\frac{4a}{a^2+4}\)
\(=\frac{a^3-1}{a^3-1}\cdot\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
Vậy \(M=\frac{4a}{a^2+4}\left(a\ne0;a\ne1\right)\)
b) \(M=\frac{4a}{a^2+4}\left(a\ne0;a\ne1\right)\)
M>0 khi 4a>0 => a>0
Kết hợp với ĐKXĐ
Vậy M>0 khi a>0 và a\(\ne\)1
c) \(M=\frac{4a}{a^2+4}\left(a\ne0;a\ne1\right)\)
\(M=\frac{4a}{a^2+4}=\frac{\left(a^2+4\right)-\left(a^2-4a+4\right)}{a^2+4}=1-\frac{\left(a-2\right)^2}{a^2+4}\)
Vì \(\frac{\left(a-2\right)^2}{a^2+4}\ge0\forall a\)nên \(1-\frac{\left(a-2\right)^2}{a^2+4}\le1\forall a\)
Dấu "=" <=> \(\frac{\left(a-2\right)^2}{a^2+4}=0\)\(\Leftrightarrow a=2\)
Vậy \(Max_M=1\)khi a=2