\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có : \(P=a^2+b^2+c^2\)

\(\Rightarrow P+2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)

\(\Rightarrow P+2=\left(a+b+c\right)^2\ge0\)

\(\Rightarrow P\ge-2\)

Vậy Min_P = -2 tại a + b + c = 0 .

Dễ thấy:

\(2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow P\ge ab+bc+ca=1\)

\(minP=1\Leftrightarrow a=b=c=\dfrac{\sqrt{3}}{3}\)

Cách khác:

Áp dụng BĐT BSC:

\(ab+bc+ca=1\)

\(\Rightarrow1=\left(ab+bc+ca\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)=\left(a^2+b^2+c^2\right)^2=P^2\)

\(\Rightarrow P\ge1\left(\text{Do }P>0\right)\)

\(minP=1\Leftrightarrow a=b=c=\dfrac{\sqrt{3}}{3}\)

Ta có BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.3=9\)

\(\Rightarrow a+b+c\ge3\)

Phân tích và áp dụng BĐT AM-GM:

\(\dfrac{1+3a}{1+b^2}=\dfrac{1}{1+b^2}+\dfrac{3a}{1+b^2}=\left(1-\dfrac{b^2}{1+b^2}\right)+\left(3a-\dfrac{3ab^2}{1+b^2}\right)\ge\left(1-\dfrac{b^2}{2b}\right)+\left(3a-\dfrac{3ab^2}{2b}\right)=\left(1-\dfrac{b}{2}\right)+\left(3a-\dfrac{3}{2}ab\right)\)

Tương tự:

\(\dfrac{1+3b}{1+c^2}\ge\left(1-\dfrac{c}{2}\right)+\left(3b-\dfrac{3}{2}bc\right)\)

\(\dfrac{1+3c}{1+a^2}\ge\left(1-\dfrac{a}{2}\right)+\left(3c-\dfrac{3}{2}ca\right)\)

Cộng các vế của các BĐT ta được:

\(P\ge3-\dfrac{1}{2}\left(a+b+c\right)+3\left(a+b+c\right)-\dfrac{3}{2}\left(ab+bc+ca\right)=3+\dfrac{5}{2}\left(a+b+c\right)-\dfrac{3}{2}.3\ge3+\dfrac{5}{2}.3-\dfrac{9}{2}=6\)

\(P=6\Leftrightarrow a=b=c=1\)

Vậy \(P_{min}=6\)

https://hoc24.vn/cau-hoi/cho-abc-0-thoa-man-abbcca3-tim-gia-tri-nho-nhat-cua-pdfrac13a1b2dfrac13b1c2dfrac13c1a2.6181078378966

Ta có: 2P=(a²+b²) + (b²+c²) + (c²+a²) 

Theo Cauchy có: 

\(2P\ge2ab+2bc+2ca=2\left(ab+bc+ca\right)=2.9\)

=> \(P\ge9\)=> P_min = 9 đạt được khi x=y=\(\sqrt{3}\)

Hoặc:

P²= (a²+b²+c²)(b²+c²+a²) 

Theo Bunhiacopxki có:

P²= (a²+b²+c²)(b²+c²+a²) \(\ge\)(ab+bc+ca)²=9²

=> P\(\ge\)9  => P_min=9

Vì \(a\ge1,b\ge1,c\ge1\)(gt) => \(\left(a-1\right)\left(b-1\right)\ge0\)<=> ab -a -b + 1 \(\ge0\)(1)

\(\left(b-1\right)\left(c-1\right)\ge0\)<=> bc - b - c + 1 \(\ge0\)(2)

\(\left(c-1\right)\left(a-1\right)\ge0\)<=> ca -c - a + 1 \(\ge0\)(3)

Cộng từng vế của (1), (2) và (3) ta được: 

ab + bc + ca -2(a +b +c) + 3 \(\ge0\)

=> \(a+b+c\le\frac{ab+bc+ca+3}{2}=\frac{9+3}{2}=6\)

Mà \(a\ge1,b\ge1,c\ge1\Rightarrow a+b+c\ge3\)=> \(3\le a+b+c\le6\)=> \(\left(a+b+c\right)^2\le36\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\le36\)

=> \(a^2+b^2+c^2\le36-2\left(ab+bc+ca\right)=36-2\times9=18\)=> P \(\le18\)

Vậy GTLN của P là 18 

Dâu "=" xảy ra khivà chỉ khi:

a =b=1, c=4 

hoặc: b=c=1, a=4

hoặc: c=a=1, b=4

\(P=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\) (BĐT Cauchy Schwarz)

\(=\dfrac{9}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{7}{ab+bc+ca}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

Ta có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) .Thế vào biểu thức

\(\Rightarrow P\ge9+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)

\(\Rightarrow P_{min}=30\) khi \(a=b=c=\dfrac{1}{3}\)