1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

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

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

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)

\(1+a^2b^2=abc\left(a+b+c\right)+a^2b^2=ab\left(ab+bc+ca+c^2\right)=ab\left(a+c\right)\left(b+c\right)\)

\(1+b^2c^2=bc\left(a+b\right)\left(a+c\right)\) ; \(1+a^2c^2=ac\left(a+b\right)\left(b+c\right)\)

\(\Rightarrow Q=\frac{c^2\left(a+b\right)^2ab\left(a+c\right)\left(b+c\right)}{bc\left(a+b\right)\left(a+c\right)ac\left(a+b\right)\left(b+c\right)}=1\)

ap dung bdt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) 

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

\(\Rightarrow P\le\frac{1}{16}\left[\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2+\left(\frac{1}{a+b}+\frac{1}{b+c}\right)^2+\left(\frac{1}{b+c}+\frac{1}{a+c}^2\right)\right]\)

\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(a+c^2\right)}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(a+b\right)\left(a+c\right)}\)\(+\frac{2}{\left(b+c\right)\left(c+a\right)}\)

ap dung \(x^2+y^2+z^2\ge xy+yz+xz\) voi a+b=x, b+c=y, c+a=z

\(16P\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)

tiếp tục áp dụng bdt ban đầu \(\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Rightarrow\frac{1}{\left(a+b\right)^2}\le4.16.\left(\frac{1}{a}+\frac{1}{b}\right)^2\)

\(\Rightarrow16P\le\frac{1}{4}.16\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)

=\(\frac{1}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\right)\)

tiep tuc ap dung bo de thu 2 ta co 

\(16P\le\frac{1}{4}.4\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)

\(\Rightarrow p\le\frac{3}{16}\)dau =khi a=b=c=1

Nguồn : mạng :V vào thống kê coi hình

\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)

\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)

Áp dụng BĐT Cosi ta có:

\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)

Từ (1)(2)(3) ta có:

\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)

Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)

Dấu "=" xảy ra <=> a=b=c=1

CHÚC BAN HỌC GIỎI

đại khái giống Ngọc thôi, sửa 1 số lỗi 

\(P=1-2\left(ab^2+bc^2+ca^2\right)-2abc\)

\(b=mid\left\{a;b;c\right\}\)\(\Rightarrow\)\(ab^2+ca^2\le a^2b+abc\)

\(\Rightarrow\)\(P\le1-2a^2b-2bc^2-4abc=1-2b\left(c+a\right)^2\le1-8\left(\frac{b+\frac{c+a}{2}+\frac{c+a}{2}}{3}\right)^3=\frac{19}{27}\)

ta có ab+bc+ca=(a+b+c)(ab+bc+ca)=(a²b+b²c+c²a)+(ab²+bc²+ca²)+3abc

=> a²+b²+c²=(a+b+c)²-2(ab+bc+ca)=1-2(ab+bc+ca)=1-2[(a²b+b²c+c²a)+(ab²+bc²+ca²)+3abc]

do đó P=2(a²b+b²c+c²a)+1-2[(a²b+b²c+c²a)+(ab²+bc²+ca²)+3abc]+4abc

=1-2(ab²+bc²+ca²)

không mất tính tổng quát giả sử a =<b=<c. suy ra

a(a-b)(b-c) >=0 => (a²-a)(b-c) >=0

=> a²b-a²c-ab²+abc >=0 => ab²+ca²=< a²b+abc

do đó ab²+bc²+ca²+abc=(ab²+ca²)+bc²+abc =< (a²b+abc)+b²c+abc=b(a+c)²

với các số dương x,y,z ta luôn có: \(x+y+z-3\sqrt[3]{xyz}=\frac{1}{2}\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)\left[\left(\sqrt[3]{x}-\sqrt[3]{y}\right)^2+\left(\sqrt[3]{y}-\sqrt[3]{z}\right)^2+\left(\sqrt[3]{z}-\sqrt[3]{x}\right)^2\right]\ge0\)

=> \(x+y+z\ge3\sqrt[3]{xyz}\Rightarrow xyz\le\left(\frac{x+y+z}{3}\right)^2\)(*)

dấu "=" xảy ra khi và chỉ khi x=y=z

áp dụng bđt (*) ta có:

\(b\left(a+c\right)^2=ab\left(\frac{a+c}{2}\right)\left(\frac{a+c}{2}\right)\le4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=4\left(\frac{a+b+c}{3}\right)^3=\frac{4}{27}\)

=> P=1-2(ab²+bc²+ca²+abc) >= 1-2b(a+c)² >= 1-2.4/₂₇=19/₂₇

vậy minP=19/₂₇ khi x=y=z=1/₃

Ta có: \(\frac{a^2+1}{c^2a^2}=\frac{1}{c^2}+\frac{1}{a^2c^2}=\frac{1}{c^2}+b^2\)

CMTT: \(\frac{b^2+1}{a^2b^2}=\frac{1}{a^2}+c^2\)

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

=> \(\frac{a^2+1}{c^2a^2}+\frac{b^2+1}{a^2b^2}+\frac{c^2+1}{b^2c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+a^2+b^2+c^2\)

Áp dụng bđt: x² + y² + z² \(\ge\)xy + yz + xz

CM đúng: <=> (x - y)² + (y - z)² + (z - x)² \(\ge\)0 (luôn đúng với mọi x,y, z)

Do đó: \(\frac{a^2+1}{c^2a^2}+\frac{b^2+1}{a^2b^2}+\frac{c^2+1}{b^2c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+ab+bc+ac=a+b+c+ab+bc+ac\)

\(=a\left(b+1\right)+b\left(c+1\right)+c\left(a+1\right)\)(đpcm)