Ta co:

\(M=\frac{9}{1-2\left(ab+bc+ca\right)}+\frac{2}{abc}=\frac{9}{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}+\frac{2}{abc}=\frac{9}{a^2+b^2+c^2}+\frac{2}{abc}\)

Ta lai co:

\(a+b+c=1\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{1}{abc}\)

\(\Rightarrow M=\frac{9}{\Sigma_{cyc}a^2}+\Sigma_{cyc}\frac{2}{ab}\ge\frac{9}{\Sigma_{cyc}a^2}+\frac{18}{\Sigma_{cyc}ab}\left(1\right)\)

\(VT_{\left(1\right)}=\frac{9}{\Sigma_{cyc}a^2}+\frac{1}{\Sigma_{cyc}ab}+\frac{1}{\Sigma_{cyc}ab}+\frac{16}{\Sigma_{cyc}ab}\ge\frac{\left(3+1+1\right)^2}{\Sigma_{cyc}a^2+2\Sigma_{cyc}ab}+\frac{16}{\frac{\left(\Sigma_{cyc}a\right)^2}{3}}=\text{ }\frac{25}{\left(\Sigma_{cyc}a\right)^2}+48=\text{ }73\)

Dau '=' xay ra khi \(\text{ }a=b=c=\frac{1}{3}\)

@my-friend 

\(M\ge\frac{9}{a^2+b^2+c^2}+\frac{36}{2\left(ab+bc+ca\right)}\ge\frac{\left(3+6\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=81\)

Dấu "=" xảy ra ra khi \(\hept{\begin{cases}\frac{3}{a^2+b^2+c^2}=\frac{6}{2\left(ab+bc+ca\right)}\\a+b+c=1\end{cases}}\Leftrightarrow a=b=c=\frac{1}{3}\)

Trước hết dễ có: \(\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\ge abc\)

\(\Rightarrow abc\le\frac{ab+bc+ca}{9}=t\) (với \(0< t=\frac{ab+bc+ca}{9}\le\frac{\left(a+b+c\right)^2}{27}=\frac{1}{27}\))

Do đó \(M\ge\frac{9}{1-18t}+\frac{2}{t}=\frac{2\left(27t-1\right)^2}{t\left(1-18t\right)}+81\ge81\forall0< t\le\frac{1}{27}\)

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

\(sigma\frac{a^2+b^2}{ab\left(a+b\right)^3}\ge sigma\frac{\frac{\left(a+b\right)^2}{2}}{\left(a+b\right)^2\left(a^3+b^3\right)}=sigma\frac{1}{2\left(a^3+b^3\right)}\ge\frac{9}{4\left(a^3+b^3+c^3\right)}=\frac{9}{4}\)

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

1)

Theo bđt Cauchy - Schwart ta có:

\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)

\(=\frac{\left(ab+bc+ca\right)^2}{abc\left(ab+bc+ca\right)+3a^2b^2c^2}\)

Đặt \(ab+bc+ca=x;abc=y\).

Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)

\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )

Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1

sai rồi nhé bạn 

làm sao mà \(x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\)lại luôn đúng

\(P=\frac{4a^2}{\sqrt{16b\left(b+15c\right)}}+\frac{4b^2}{\sqrt{16c\left(c+15a\right)}}+\frac{4c^2}{\sqrt{16a\left(a+15c\right)}}\)

\(\Rightarrow P\ge\frac{8a^2}{17b+15c}+\frac{8b^2}{17c+17a}+\frac{8c^2}{17a+15b}\)

\(\Rightarrow P\ge\frac{8\left(a+b+c\right)^2}{32\left(a+b+c\right)}=\frac{a+b+c}{4}\ge\frac{\sqrt{3\left(ab+bc+ca\right)}}{4}=\frac{\sqrt{3}}{4}\)

\(P_{min}=\frac{\sqrt{3}}{4}\) khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Áp dụng BĐT Cauchy-schwarz ta có:

\(P=\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\ge\frac{9}{3+ab+bc+ca}\ge\frac{9}{3+12}=\frac{3}{5}\)

Dấu " = " xảy ra <=> a=b=c=2

Áp dụng BĐT AM-GM,ta có:

\(P\ge3\sqrt[3]{\frac{1}{\left(1+ab\right)\left(1+bc\right)\left(1+ca\right)}}=\frac{3}{\sqrt[3]{\left(1+ab\right)\left(1+bc\right)\left(1+ca\right)}}\)

\(\ge\frac{3}{\frac{\left(3+ab+bc+ca\right)}{3}}=\frac{9}{3+ab+bc+ca}\)

Ta có BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)

Áp dụng vào,ta có: \(P\ge\frac{9}{3+ab+bc+ca}\ge\frac{9}{3+a^2+b^2+c^2}=\frac{9}{15}=\frac{3}{5}\)

Thật ra cũng chả khác kudo tí nào,nên đừng bắt bẻ nhá.Chỉ khác cách trình bày thôi:v

@NTMH

Ta chứng minh \(P\ge\frac{9}{2}\). Ta đã có: \(\frac{a^3+b^3+c^3}{2abc}\ge\frac{3abc}{2abc}=\frac{3}{2}\)

Vậy cần chứng minh \(\frac{a^{2}+b^{2}}{c^{2}+ab}+\frac{b^{2}+c^{2}} {a^{2}+bc}+\frac{c^{2}+a^{2}}{b^{2}+ac}\geq 3\)

\(\Leftrightarrow a^{2}(\frac{1}{c^{2}+ab}+\frac{1}{b^{2}+ac)}+b^{2}(\frac{1}{c^{2}+ab}+\frac{1}{a^{2}+bc})+c^{2}( \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ac})\)

\(\geq \frac{4a^{2}}{(a+b)(b+c)}+\frac{4b^{2}}{(c+a)(c+b) }+\frac{4c^{2}}{(a+b)(a+c)}\)

\(\geq \frac{4(a+b+c)^{2}}{(a+b)(b+c)+(c+a)(c+b)+(a+c)(a+ b)}\geq 3\)

BĐT đã được chứng minh

Vậy ta có \(P_{min}=\frac{9}{2}\) khi \(a=b=c\)