2. a) \(\left\{{}\begin{matrix}x,y,z>1\\x+y+z=xyz\end{matrix}\right.\) Tìm min \(P=\frac{x-1}{y^2}+\frac{y-1}{z^2}+\frac{z-1}{x^2}\)
b) \(a,b,c>0.Cmr:\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
c) \(\left\{{}\begin{matrix}x,y,z\ge0\\x^2+y^2+z^2=2\end{matrix}\right.\) Tìm max \(P=\frac{x^2}{x^2+yz+x+1}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}\)
d) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{ab+3c}+\frac{b}{bc+3a}+\frac{c}{ca+3b}\ge\frac{3}{4}\)
\(A=\frac{a}{ab+c\left(a+b+c\right)}+\frac{b}{bc+a\left(a+b+c\right)}+\frac{c}{ca+b\left(a+b+c\right)}\)
\(=\frac{a}{\left(b+c\right)\left(a+c\right)}+\frac{b}{\left(a+b\right)\left(a+c\right)}+\frac{c}{\left(a+b\right)\left(c+b\right)}\)
Áp dụng bđt AM-GM ta có
\(A=\frac{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\ge27.\frac{a^2+b^2+c^2+ab+bc+ca}{8\left(a+b+c\right)^3}\)\(=\frac{a^2+b^2+c^2+ab+bc+ca}{8}\)
\(=\frac{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}{8}\)\(\ge\frac{9-\frac{\left(a+b+c\right)^2}{3}}{8}=\frac{9-3}{8}=\frac{3}{4}\)
Dấu "=" xảy ra khi a=b=c=1
b) Mạnh hơn, và dễ dàng hơn là:
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{\sum c\left(a-b\right)^2}{abc}\)
Nó tương đương với: \({\frac {{a}^{2}}{{b}^{2}}}+{\frac {{b}^{2}}{{c}^{2}}}+{\frac {{c}^{2} }{{a}^{2}}}+3-2\,{\frac {a}{b}}-2\,{\frac {b}{c}}-2\,{\frac {c}{a}} \geqq 0\)
Là hiển nhiên vì \(\frac{a^2}{b^2}+1\ge\frac{2a}{b}\)
Đơn giản:))
a) Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow ab+bc+ca=1;0< a,b,c< 1\)
Cần chứng minh: \(P=\sum\frac{\frac{1}{a}-1}{\frac{1}{b^2}}=\sum\frac{b^2-ab^2}{a}\ge\sqrt{3}-1\)
Hay là: \(\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)\sqrt{ab+bc+ca}\ge\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)+a^2+b^2+c^2\)
\(\Leftrightarrow\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)^2\left(ab+bc+ca\right)\ge\) \(\Big[ (\sqrt{3} -1) (ab+bc+ca) +a^2+b^2+c^2\Big]^2\)
Giả sử \(c=\min\{a,b,c\}\) và đặt \(a=c+u, \, b=c+v \, (u,\, v \geq 0)\)
Nếu mình không nhìn nhầm, sau khi rút gọn, nhóm lại theo biến c, bạn nhận được một cái gì đó gọi là hiển nhiên 
Chúc may mắn, mình mới rút gọn thử thì thấy có vẻ hiển nhiên thật :))
a/ Một cách đơn giản hơn:
\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
\(P=\frac{x-\frac{1}{2}+y-\frac{1}{2}}{y^2}+\frac{y-\frac{1}{2}+z-\frac{1}{2}}{z^2}+\frac{z-\frac{1}{2}+x-\frac{1}{2}}{x^2}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(P=\left(x-\frac{1}{2}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\left(y-\frac{1}{2}\right)\left(\frac{1}{y^2}+\frac{1}{z^2}\right)+\left(z-\frac{1}{2}\right)\left(\frac{1}{x^2}+\frac{1}{z^2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(P\ge\frac{2}{xy}\left(x-\frac{1}{2}\right)+\frac{2}{yz}\left(y-\frac{1}{2}\right)+\frac{2}{zx}\left(z-\frac{1}{2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(P\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\)
\(P\ge\sqrt{3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}-1=\sqrt{3}-1\)
\(P_{min}=\sqrt{3}-1\) khi \(x=y=z=\sqrt{3}\)
b/ \(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(ab+bc+ca\right)\ge\left(a+b+c\right)^2\) (1)
\(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)^2\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{a+b+c}{abc}\right)\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\) (2)
(1);(2) \(\Rightarrow\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\left(\frac{ab+bc+ca}{abc}\right)\left(a+b+c\right)\ge\left(a+b+c\right)^2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
c/ \(2+2yz=x^2+y^2+z^2+2yz=x^2+\left(y+z\right)^2\ge2x\left(y+z\right)\)
\(\Rightarrow x\left(y+z\right)\le1+yz\)
Mặt khác cũng có: \(2+2yz=x^2+\left(y+z\right)^2\ge\frac{1}{2}\left(x+y+z\right)^2\Rightarrow1+yz\ge\frac{\left(x+y+z\right)^2}{4}\)
\(P\le\frac{x^2}{x^2+x+x\left(y+z\right)}+\frac{y+z}{x+y+z+1}-\frac{\left(x+y+z\right)^2}{36}\)
\(P\le\frac{x+y+z}{x+y+z+1}-\frac{\left(x+y+z\right)^2}{36}\)
Đặt \(x+y+z=t\Rightarrow P\le\frac{t}{t+1}-\frac{t^2}{36}=\frac{36t-t^3-t^2}{36\left(t+1\right)}\)
\(P\le\frac{-t^3-t^2+16t-20+20\left(t+1\right)}{36\left(t+1\right)}=\frac{-\left(t-2\right)^2\left(t+5\right)}{36\left(t+1\right)}+\frac{5}{9}\le\frac{5}{9}\)
\(\Rightarrow P_{max}=\frac{5}{9}\) khi \(t=2\) hay \(\left(x;y;z\right)=\left(1;1;0\right);\left(1;0;1\right)\)
d/ \(VT=\frac{a}{\left(a+c\right)\left(b+c\right)}+\frac{b}{\left(a+b\right)\left(a+c\right)}+\frac{c}{\left(a+b\right)\left(b+c\right)}\)
\(VT=\frac{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{a^2+b^2+c^2+ab+bc+ca}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(VT\ge\frac{\frac{1}{2}\left(a+b+c\right)^2+\frac{1}{2}\left(a^2+b^2+c^2\right)}{\left(\frac{2a+2b+2c}{3}\right)^3}\ge\frac{\frac{1}{2}\left(a+b+c\right)^2+\frac{1}{6}\left(a+b+c\right)^2}{\frac{8}{27}\left(a+b+c\right)^3}=\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cách 3 cho câu b, khá độc đáo:
Đặt \(\left(a;b;c\right)\rightarrow\left(a^2;b^2;c^2\right)\) (nên đặt x2.. nhưng mình đặt vậy để sẽ nói sau)
Đưa bất đẳng thức về: \(\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^2\ge\left(a^2+b^2+c^2\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
Áp dụng bổ đề với \(x=\frac{a}{b};y=\frac{b}{c};z=\frac{c}{a}\)
Có: \(VT\ge\left[\frac{3}{2}\sum\left(x+\frac{1}{x}\right)-6\right]^2\) (viết x, y, z lại như bổ đề trong link cho dễ xem :v)
\(=\left[\frac{3}{2}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right)-6\right]^2\ge VP\)
Hoán vị trở thành đối xứng, đơn giản chưa:v
