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

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

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

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

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

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

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

Dấu "=" xảy ra khi \(a=b=c=1\)

\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)

Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)

Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)

\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)

\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)

\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)

\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)

\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)

\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)

\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)

\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)

Từ giả thiết \(2\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{2}{3}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)

Đặt \(Q=\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}\)

\(=\dfrac{a^2b^2c^2+\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)+1}{abc}\)

\(\ge\dfrac{a^2b^2c^2+3\sqrt[3]{\left(a^2b^2c^2\right)^2}+3\sqrt[3]{a^2b^2c^2}+1}{abc}=\dfrac{\left(\sqrt[3]{a^2b^2c^2}+1\right)^3}{abc}\)

\(=\left(\dfrac{\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3=\left(\sqrt[3]{abc}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)

\(=\left(\sqrt[3]{abc}+\dfrac{4}{9\sqrt[3]{abc}}+\dfrac{5}{9\sqrt[3]{abc}}\right)^3\ge\left(2\sqrt[]{\dfrac{4\sqrt[3]{abc}}{9\sqrt[3]{abc}}}+\dfrac{5}{9.\dfrac{2}{3}}\right)^3=\dfrac{2197}{216}\)

\(\Rightarrow P\ge3\sqrt[3]{\dfrac{2197}{216}}=\dfrac{13}{2}\)

ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé

\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)

\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)

<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)

\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)

áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)

\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)

\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)

<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)

dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)

\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)

\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)

Từ đó:

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

\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)

Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)

Cộng vế:

\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)

\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)

\(\sqrt{ab}+\sqrt{4b.c}+2\left(a+c\right)\le\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(4b+c\right)+2\left(a+c\right)=\dfrac{5}{2}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{2}{5}\left(\dfrac{1}{a+b+c}-\dfrac{1}{\sqrt{a+b+c}}\right)=\dfrac{2}{5}\left(\dfrac{1}{\sqrt{a+b+c}}-\dfrac{1}{2}\right)^2-\dfrac{1}{10}\ge-\dfrac{1}{10}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a+b+c=4\\a=b=\dfrac{c}{4}\end{matrix}\right.\) em tự giải ra a;b;c

Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)

Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì  \(\left(a-b=b-c\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)

\(P=2\Sigma a+\Sigma\dfrac{1}{a}=\Sigma a+\Sigma a+\Sigma\dfrac{1}{a}\ge3.\sqrt[3]{\left(\Sigma a\right)^2.\Sigma\dfrac{1}{a}}\)

\(Q=\left(\Sigma a\right)^2.\Sigma\dfrac{1}{a}=\left(3+2\Sigma ab\right).\Sigma\dfrac{1}{a}=3\Sigma\dfrac{1}{a}+4\Sigma a+2\Sigma\dfrac{ab}{c}\ge3\Sigma\dfrac{1}{a}+6\Sigma a=3\left(\Sigma\dfrac{1}{a}+2\Sigma a\right)=3P\)\(\Rightarrow\)\(P\ge3\sqrt[3]{3P}\)   \(\Leftrightarrow P^3\ge81P\Leftrightarrow P^2\ge81\left(P>0\right)\Leftrightarrow P\ge9\)

" = " \(\Leftrightarrow a=b=c=1\)

Vì $\large a,b,c \in\mathbb{N^*}$ và $\large a^2+b^2+c^2=3\Rightarrow \left\{\begin{matrix} a<\sqrt{3} & \\ b<\sqrt{3} & \\ c<\sqrt{3} & \end{matrix}\right.$

Ta chứng minh bất đẳng thức phụ sau: 

Với $0 <x<\sqrt{3}$ thì $2x+\frac{1}{x} \ge x^2.\frac{1}{2}+\frac{5}{2}(*)$

Thật vậy $(*)$ $\large \Leftrightarrow (x-2)(x-1)^2 \le0$

Do $\large x<\sqrt{3}\Leftrightarrow x<2\Leftrightarrow (x-2)(x-1)^2<0$ (Luôn đúng)

Do đó bất đẳng thức được chứng minh 

Dấu $"="$ xảy ra khi $x=1$

Trở lại bài toán: 

Áp dụng BĐT $(*)$ ta được:

$\large 2a+\frac{1}{a}+2b+\frac{1}{b}+2c+\frac{1}{c}\ge\frac{1}{2}(a^2+b^2+c^2)+\frac{15}{2}=9$

Do $a^2+b^2+c^2=3$

Vậy $GTNN=9$

Dấu $"="$ xảy ra khi: $a=b=c=1$