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\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
\(A=\frac{a}{ab+c\left(a+b+c\right)}+\frac{b}{bc+a\left(a+b+c\right)}+\frac{c}{ca+b\left(a+b+c\right)}\)
\(=\frac{a}{\left(b+c\right)\left(a+c\right)}+\frac{b}{\left(a+b\right)\left(a+c\right)}+\frac{c}{\left(a+b\right)\left(c+b\right)}\)
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
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 \(x=y=z=\sqrt{3}\)
34, Quảng Ninh
Cho x;y;z > 0 thỏa mãn x + y + z < 1
Tìm GTNN của biểu thức \(P=\frac{1}{x^2+y^2+z^2}+\frac{2019}{xy+yz+zx}\)
Ta có bđt sau : \(\frac{m^2}{a}+\frac{n^2}{b}\ge\frac{\left(m+n\right)^2}{a+b}\left(a;b>0\right)\)
Áp dụng ta được \(P=\frac{1}{x^2+y^2+z^2}+\frac{2019}{xy+yz+zx}\)
\(=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}+\frac{2017}{xy+yz+zx}\)
\(\ge\frac{\left(1+2\right)^2}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}+\frac{2017}{\frac{\left(x+y+z\right)^2}{3}}\)
\(=\frac{9}{\left(x+y+z\right)^2}+\frac{6051}{\left(x+y+z\right)^2}\)
\(=\frac{6060}{\left(x+y+z\right)^2}\ge\frac{6060}{1}=6060\)
Dấu "=" tại x = y = z = 1/3
39, Chuyên Hưng Yên
Với x;y là các số thực thỏa mãn \(\left(x+2\right)\left(y-1\right)=\frac{9}{4}\)
Tìm \(A_{min}=\sqrt{x^4+4x^3+6x^2+4x+2}+\sqrt{y^4-8y^3+24y^2-32y+17}\)
Ta có \(A=\sqrt{x^4+4x^3+6x^2+4x+2}+\sqrt{y^4-8y^3+24y^2-32y+17}\)
\(=\sqrt{\left(x+1\right)^4+1}+\sqrt{\left(y-2\right)^4+1}\)
Đặt \(\hept{\begin{cases}x+1=a\\y-2=b\end{cases}}\)
Thì \(A=\sqrt{a^4+1}+\sqrt{b^4+1}\)và giả thiết đã cho trở thành \(\left(a+1\right)\left(b+1\right)=\frac{9}{2}\)
Ta có bất đẳng thức \(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\)(1)
Thật vậy
\(\left(1\right)\Leftrightarrow x^2+y^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}+z^2+t^2\ge x^2+2xz+z^2+y^2+2yt+t^2\)
\(\Leftrightarrow\sqrt{x^2z^2+x^2t^2+y^2z^2+y^2t^2}\ge xz+yt\)
*Nếu xz + yt < 0 thì bđt luôn đúng
*Nếu xz + yt > 0 thì bđt tương đương với
\(x^2z^2+x^2t^2+y^2z^2+y^2t^2\ge x^2z^2+2xyzt+y^2t^2\)
\(\Leftrightarrow x^2t^2-2xyzt+y^2z^2\ge0\)
\(\Leftrightarrow\left(xt-yz\right)^2\ge0\)(Luôn đúng)
Vậy bđt (1) được chứng minh
Áp dụng (1) ta được \(A=\sqrt{a^4+1}+\sqrt{b^4+1}\ge\sqrt{\left(a^2+b^2\right)^2+\left(1+1\right)^2}\)
\(=\sqrt{\left(a^2+b^2\right)^2+4}\)
Ta có \(\left(a+1\right)\left(b+1\right)=\frac{9}{4}\)
\(\Leftrightarrow ab+a+b+1=\frac{9}{4}\)
\(\Leftrightarrow ab+a+b=\frac{5}{4}\)
Áp dụng bđt Cô-si có \(a^2+b^2\ge2ab\)
\(2\left(a^2+\frac{1}{4}\right)\ge2a\)
\(2\left(b^2+\frac{1}{4}\right)\ge2b\)
Cộng 3 vế vào được
\(3\left(a^2+b^2\right)+1\ge2\left(ab+a+b\right)=\frac{5}{2}\)
\(\Rightarrow a^2+b^2\ge\frac{1}{2}\)
Khi đó \(A\ge\sqrt{\left(a^2+b^2\right)^2+4}\ge\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{3}\)
Dấu ''=" tại \(\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x+1=\frac{1}{2}\\y-2=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{5}{2}\end{cases}}\)