\(b^4+c^4-bc\left(b^2+c^2\right)=\left(b^2+bc+c^2\right)\left(b-c\right)^2\)

\(\Rightarrow b^4+c^4\ge bc\left(b^2+c^2\right)\)

Tương tự\(\Rightarrow\Sigma_{cyc}\frac{a}{a+b^4+c^4}\le\Sigma_{cyc}\frac{a}{a+bc\left(b^2+c^2\right)}=\Sigma_{cyc}\frac{a}{bc\left(a^2+b^2+c^2\right)}=\frac{1}{a^2+b^2+c^2}\Sigma_{cyc}\frac{a}{bc}\)

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

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

oke rồi he

@Nub :v

Áp dụng Bunhiacopski ta dễ có:

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

Tương tự:

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

Cộng lại:

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

Ta đi chứng minh:

\(\frac{a^4+b^4+c^4+2a+2b+2c}{\left(a^2+b^2+c^2\right)^2}\le1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)

Cái này luôn  đúng theo Cauchy

Đẳng thức xảy ra tại a=b=c=1

\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có : 

\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)

\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)

\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)

Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)

PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))

nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm

Tính ra a+b+c<=4 nhé (dùng Bu-nhi-a cop-xki)

Phần còn lại tự xử nhé)

là sao zay ah???

Áp dụng AM - GM 

\(P=\frac{1}{\sqrt{a^2+b^2}}+\frac{1}{\sqrt{b^2+c^2}}+\frac{1}{\sqrt{c^2+a^2}}\ge\frac{1}{\sqrt{2ab}}+\frac{1}{\sqrt{2bc}}+\frac{1}{\sqrt{2ca}}\)

\(abc=a+b+c+2\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)+\left(b+1\right)\left(c+1\right)+\left(c+1\right)\left(a+1\right)\ge\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

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

Với mọi số thực x,y,z ta có ngay:

\(\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)

\(\Leftrightarrow\frac{1}{1+\frac{y+z}{x}}+\frac{1}{1+\frac{z+x}{y}}+\frac{1}{1+\frac{x+y}{z}}=1\)

Khi đó ta có thể đặt được \(\left(a;b;c\right)\rightarrow\left(\frac{y+z}{x};\frac{z+x}{y};\frac{x+y}{z}\right)\) 

Thay vào thì dễ có:

\(\sqrt{\frac{xy}{\left(y+z\right)\left(z+x\right)}}+\sqrt{\frac{yz}{\left(z+x\right)\left(x+y\right)}}+\sqrt{\frac{zx}{\left(z+y\right)\left(x+y\right)}}\)

\(\le\frac{1}{2}\Sigma\left(\frac{x}{x+z}+\frac{z}{x+z}\right)=\frac{3}{2}\)

Vậy ...........................

ÁP dụng BĐT AM-Gm  ta có: 

\(Σ\frac{a^2}{\left(ab+2\right)\left(2ab+1\right)}\ge\frac{4}{9}\cdotΣ\frac{a^2}{\left(ab+1\right)^2}\)

ĐẶt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\) thì cần cm

\(Σ\frac{a^2}{\left(ab+1\right)^2}=Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{3}{4}\)

\(Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\left(\frac{xz}{y\left(x+z\right)}\right)^2\)

Theo C-S \(Σ\frac{xz}{y\left(x+z\right)}=\frac{\left(xz\right)^2}{xyz\left(x+z\right)}\ge\frac{\left(Σxy\right)^2}{2xy\left(Σx\right)}\ge\frac{3}{2}\)

\(\frac{1}{3}\cdot\left(Σ\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}\)

Đúng hay ta có ĐPCM xyar ra khi a=b=c=1

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 \(a=b=c=\frac{1}{3}\)

1a

\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)

\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Vay \(A_{min}=\frac{161}{16}\)

1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)

\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)

Bài 2 Dùng Cauchy-Schwarz dạng Engel là ra:D

Bài 3:Đừng vội dùng Cauchy-Schwarz dạng Engel ngay kẻo bị phức tạp:v Thay vào đó hãy khai triển nó ra:

\(A=x^2+y^2+2\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{x^2}+\frac{1}{y^2}\)

\(\ge4+2.2+\frac{4}{x^2+y^2}=4+4+1=9\)

Đẳng thức xảy ra khi \(x=y=\sqrt{2}\)

Bài 4: Dùng Cauchy or Bunhiacopxki là ok!