theo bđt cauchy-schwarz ta có \(P\ge\frac{\left(1+1+1\right)^2}{3+2\left(a^3+b^3+c^3\right)}=\frac{9}{3+2\left(a^3+b^3+c^3\right)}\)
Mà\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3=3abc}\)\(\Rightarrow P\ge\frac{9}{3+2\cdot3abc}=\frac{9}{3+6}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vậy \(P_{max}=1\Leftrightarrow a=b=c=1\)
Sorry mình viết nhầm nha \(3\sqrt[3]{a^3b^3c^3}=3abc\)mới đúng nha
Nguyễn Gia Huy làm lộn hết dấu rồi??GTLN???
Ta có \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Rightarrow1+a^3+b^3=abc+a^3+b^3\ge ab\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{1+a^3+b^3}\le\frac{1}{ab\left(a+b+c\right)}\)
Tương tự \(\frac{1}{1+b^3+c^3}\le\frac{1}{bc\left(a+b+c\right)}\)
\(\frac{1}{1+c^3+â^3}\le\frac{1}{ca\left(a+b+c\right)}\)
\(\Rightarrow P\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}=\frac{a+b+c}{a+b+c}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Dễ dàng chứng minh \(a^3+b^3\ge ab\left(a+b\right)\)
mà \(1=abc\Rightarrow ab=\frac{1}{c}\)
\(\Rightarrow1+a^3+b^3\ge\frac{1}{c}\left(a+b\right)+1=\frac{a+b}{c}+1=\frac{a+b+c}{c}\)
\(\Rightarrow\frac{1}{1+a^3+b^3}\le\frac{c}{a+b+c}\)
Tương tự :\(\frac{1}{1+b^3+c^3}\le\frac{a}{a+b+c}\)
\(\frac{1}{1+c^3+a^3}\le\frac{b}{a+b+c}\)
\(\Rightarrow P\le\frac{a+b+c}{a+b+c}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vậy \(P_{max}=1\Leftrightarrow a=b=c=1\)
Ta có bđt phụ quen thuộc sau \(x^3+y^3\ge xy\left(x+y\right)\)(*)
Chứng minh :\(bđt< =>\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)
\(< =>\left(x+y\right)\left(x^2-2xy+y^2\right)\ge0< =>\left(x+y\right)\left(x-y\right)^2\ge0\)(đúng)
Sử dụng bđt (*) vào bài toán \(P=\frac{1}{1+a^3+b^3}+\frac{1}{1+b^3+c^3}+\frac{1}{1+c^3+a^3}\le\frac{1}{1+ab\left(a+b\right)}+\frac{1}{1+bc\left(b+c\right)}+\frac{1}{1+ca\left(c+a\right)}\)
\(< =>P\le\frac{1}{abc+ab\left(a+b\right)}+\frac{1}{abc+bc\left(b+c\right)}+\frac{1}{abc+ca\left(c+a\right)}\)\(< =>P\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)\(< =>P\le\frac{c}{abc\left(a+b+c\right)}+\frac{a}{abc\left(a+b+c\right)}+\frac{b}{abc\left(a+b+c\right)}\)
\(< =>P\le\frac{c}{a+b+c}+\frac{a}{a+b+c}+\frac{b}{a+b+c}=\frac{a+b+c}{a+b+c}=1\left(Do:abc=1\right)\)
Dấu "=" xảy ra khi và chỉ khi a = b = c = 1
Vậy GTLN của P = 1 đạt được khi a = b = c = 1