ta có \(P=a^3+b^3+c^3+3abc=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)+3abc\)

              \(=1-3\left(1-a\right)\left(1-b\right)\left(1-a\right)+3abc\)

nhân tung ra và rút gọn thì \(P=1-3\left(ab+bc+ca\right)+6abc=1-3\left(ab+bc+ca-2abc\right)\)

vì \(b+c>a\Rightarrow a+b+c\ge2a\Rightarrow2a-1< 0\)

tương tự với mấy cái kia nhân vaò và ta có 

\(\left(2a-1\right)\left(2b-1\right)\left(2c-1\right)< 0\)\(\Leftrightarrow8abc-4\left(ab+bc+ca\right)+2\left(a+b+c\right)-1< 0\)

=> \(1< 4\left(ab+bc+ca\right)-8abc\Rightarrow\frac{1}{4}< \left(ab+bc+ca-2abc\right)\)

=> \(\Rightarrow-3\left(ab+bc+ca-2abc\right)< -\frac{3}{4}\)

=> \(1-3\left(ab+bc+ca-2abc\right)< \frac{1}{4}\) => p<1/4

B) ta có \(\left(a+b-c\right)\left(a-b+c\right)\left(b+c-a\right)=\sqrt{\left[b^2-\left(a-c\right)^2\right]\left[a^2-\left(b-c\right)^2\right]\left[c^2-\left(a-b\right)^2\right]}< abc\)

=> \(\left(1-2a\right)\left(1-2b\right)\left(1-2c\right)< abc\)

=> \(4\left(ab+bc+ca-2abc\right)\le abc+1\le\left(\frac{a+b+c}{3}\right)^3+1=\frac{28}{27}\)

=> \(ab+bc+ca-abc\le\frac{7}{27}\)

=> \(P\ge1-3.\frac{7}{27}=\frac{2}{9}\)

Ta có a+b+c=1;a;b;c>0 nên

P=a³+b³+c³+3abc

=(a+b+c)³-3(a+b)(b+c)(c+a)+3abc

=1-3abc-3∑ab(a+b)

=1-3abc-3∑ab(1-c)

=1-3(ab+bc+ca)+6abc

Vì a;b;c là 3 cạnh của một tam giác nên

b+c>a=>a+b+c>2a=>2a<1. Tương tự 2b<1;2c<1

Nên (2a-1)(2b-1)(2c-1)<0

<=> 8abc-4(ab+bc+ca)+2(a+b+c)-1<0

=>4[ab+bc+ca-2abc]>1

=>P<1/4

Ta có:

(a+b-c)(b+c-a)(c+a-b)=

\(\sqrt{\left[b^2-\left(a-c\right)^2\right].\left[a^2-\left(b-c\right)^2\right].\left[c^2-\left(a-b\right)^2\right]}\)≤abc

=>(1-2a)(1-2b)(1-2c)≤abc

=>4[ab+bc+ca-2abc]≤abc+1≤\(\left(\frac{a+b+c}{3}\right)^3+1=\frac{28}{27}\)

=>P≥1-3.\(\frac{28}{4.27}=\frac{2}{9}\)

Dấu = xảy ra khi a=b=c=\(\frac{1}{3}\)

trời mãi ms xong