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

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

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

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

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

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

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

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\dfrac{a^5}{b^3+c^2}+\dfrac{b^3+c^2}{4}+\dfrac{a^4}{2}\ge3\sqrt[3]{\dfrac{a^9.\left(b^3+c^2\right)}{8\left(b^3+c^2\right)}}=\dfrac{3a^3}{2}\)

Tương tự và cộng lại:

\(\Rightarrow M-\dfrac{a^4+b^4+c^4}{2}+\dfrac{a^3+b^3+c^3}{4}+\dfrac{a^2+b^2+c^2}{4}\ge\dfrac{3}{2}\left(a^3+b^3+c^3\right)\)

\(\Rightarrow M\ge\dfrac{a^4+b^4+c^4}{2}+\dfrac{5}{4}\left(a^3+b^3+c^3\right)-\dfrac{3}{4}\)

Mặt khác ta có:

\(\dfrac{1}{2}\left(a^4+b^4+c^4\right)\ge\dfrac{1}{6}\left(a^2+b^2+c^2\right)^2=\dfrac{3}{2}\)

\(\left(a^3+a^3+1\right)+\left(b^3+b^3+1\right)+\left(c^3+c^3+1\right)\ge3\left(a^2+b^2+c^2\right)=9\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge9\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{15}{4}-\dfrac{3}{4}=...\)

Lời giải:

Đặt $a+b+c=p; ab+bc+ac=q=1; abc=r$

$p,r\geq 0$

Áp dụng BĐT AM-GM: $p^2\geq 3q=3\Rightarrow p\geq \sqrt{3}$

$a,b,c\leq 1\Leftrightarrow (a-1)(b-1)(c-1)\leq 0$

$\Leftrightarrow p+r\leq 2\Rightarrow p\leq 2$

$P=\frac{(a+b+c)^2-2(ab+bc+ac)+3}{a+b+c-abc}=\frac{(a+b+c)^2+1}{a+b+c-abc}=\frac{p^2+1}{p-r}$

Ta sẽ cm $P\geq \frac{5}{2}$ hay $P_{\min}=\frac{5}{2}$

$\Leftrightarrow \frac{p^2+1}{p-r}\geq \frac{5}{2}$

$\Leftrightarrow 2p^2-5p+2+5r\geq 0(*)$

---------------------------

Thật vậy:

Áp dụng BĐT Schur thì:

$p^3+9r\geq 4p\Rightarrow 5r\geq \frac{20}{9}p-\frac{5}{9}p^3$

Khi đó:

$2p^2-5p+2+5r\geq 2p^2-5p+2+\frac{20}{9}p-\frac{5}{9}p^3=\frac{1}{9}(2-p)(5p^2-8p+9)\geq 0$ do $p\leq 2$ và $p\geq \sqrt{3}$

$\Rightarrow (*)$ được CM

$\Rightarrow P_{\min}=\frac{5}{2}$

Dấu "=" xảy ra khi $(a,b,c)=(1,1,0)$ và hoán vị

cái cuối là \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\)  nha

\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế:

\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)

Đơn giản là kiên nhẫn tính toán và tách biểu thức:

\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)

Sau đó Cô-si cho từng ngoặc là được

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

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

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

Áp dụng BĐT Holder:

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

Mặt khác:

\(\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\dfrac{3}{2}\left(a^2b^2+b^2c^2+c^2a^2+abc\left(a+b+c\right)\right)\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{3}{4}\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\)

\(\Rightarrow\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\ge\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}\)

\(\Rightarrow P\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}+\dfrac{4}{\sqrt{a^2+b^2+c^2+1}}\)

Đặt \(\sqrt{\dfrac{a^2+b^2+c^2}{3}}=x>0\)

\(\Rightarrow P\ge\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\)

Ta sẽ chứng minh \(P\ge\dfrac{7}{2}\)

Thật vậy, với \(x\ge\dfrac{7}{3}\Rightarrow P>\dfrac{3x}{2}\ge\dfrac{7}{2}\) (đúng)

Với \(0< x\le\dfrac{7}{3}\) ta cần chứng minh:

\(\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7}{2}\Leftrightarrow\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7-3x}{2}\)

\(\Leftrightarrow64\ge\left(7-3x\right)^2\left(3x^2+1\right)\)

\(\Leftrightarrow3\left(x-1\right)^2\left(-9x^2+24x+5\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)^2\left[3x\left(7-3x\right)+3x+5\right]\ge0\) (đúng)

Vậy \(P_{min}=\dfrac{7}{2}\) khi \(x=1\) hay \(a=b=c=1\)