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\(xy+yz+zx\le3xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)
\(P=\frac{1}{\sqrt{x^2+y^2+x^2+xy}}+\frac{1}{\sqrt{y^2+z^2+y^2+yz}}+\frac{1}{\sqrt{z^2+x^2+z^2+zx}}\)
\(P\le\frac{1}{\sqrt{x^2+3xy}}+\frac{1}{\sqrt{y^2+3yz}}+\frac{1}{\sqrt{z^2+3zx}}=\frac{4}{2\sqrt{4x\left(x+3y\right)}}+\frac{4}{2\sqrt{4y\left(y+3z\right)}}+\frac{1}{2\sqrt{4z\left(z+3x\right)}}\)
\(P\le4\left(\frac{1}{4x+x+3y}+\frac{1}{4y+y+3z}+\frac{1}{4z+z+3x}\right)=4\left(\frac{1}{5x+3y}+\frac{1}{5y+3z}+\frac{1}{5z+3x}\right)\)
\(P\le\frac{4}{64}\left(\frac{5}{x}+\frac{3}{y}+\frac{5}{y}+\frac{3}{z}+\frac{5}{z}+\frac{3}{x}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{3}{2}\)
\(P_{max}=\frac{3}{2}\) khi \(x=y=z=1\)
giờ nhân cả tử và mẫu mỗi phân thức vs mỗi tử của nó rồi sử dụng BDT bunhiacopxki là ra thôi bn
\(\frac{x^2}{x^3-xyz+2013x}+\frac{y^2}{y^3-xyz+2013y}+\frac{z^2}{z^3-xyz+2013z}\)
\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+3.\left(xy+yz+zx\right)\left(x+y+z\right)}\)
\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx+3xy+3yz+3zx\right)}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x+y+z\right)^2}=\frac{1}{x+y+z}\)
a) ĐKXĐ: \(x;y>0\)
Ta có:\(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\)
\(\Rightarrow\frac{4y}{4xy}+\frac{4x}{4xy}=\frac{xy}{4xy}\)
\(\Rightarrow4x+4y-xy=0\)
\(\Rightarrow x\left(4-y\right)=-4y\)
\(\Rightarrow x=\frac{-4y}{4-y}=\frac{-4\left(y-4\right)-16}{-\left(y-4\right)}\)
\(\Rightarrow x=4-\frac{16}{4-y}\)
Để x nguyên dương =>\(\hept{\begin{cases}\frac{16}{4-y}< 0\\\left(4-y\right)\inƯ\left(16\right)\end{cases}}\)
\(\Rightarrow4-y\in\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)
Tìm nốt y và thay vào tìm ra x
a/ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\)
Không mất tính tổng quát giả sử: \(x\ge y\)
\(\frac{1}{4}=\frac{1}{x}+\frac{1}{y}\le\frac{2}{y}\)
\(\Leftrightarrow0< y\le8\)
\(\Rightarrow y=\left\{1;2;3;4;5;6;7;8\right\}\)làm nốt
trước tiên ta phải cm: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(#\right)\left(\forall a,b,c\in R;x,y,z>0\right)\)
dấu = xảy ra khi zà chỉ khi\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
thật zậy , zới \(a,b\in R;x,y>0\)ta có \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(##\right)\left(a,b\in R;x,y>0\right)\)
\(\Leftrightarrow\left(a^2y+b^2x\right)\left(x+y\right)\ge xy\left(a+b\right)^2\Leftrightarrow\left(bx-ay\right)^2\ge0\)( luôn đúng )
dấu = xảy ra khi zà chỉ khi\(\frac{a}{x}=\frac{b}{y}\)
* áp dụng bất đẳng thức (##) ta được
\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
dấu = xảy ra khi zà chỉ khi \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)\
* áp dụng bất đẳng thức (#) ta có
vt = \(\frac{x^2}{x\left(x^2-yz+2010\right)}+\frac{y^2}{y\left(y^2-xz+2010\right)}+\frac{z^2}{z\left(z^2-xy+2010\right)}\)
=\(\frac{x^2}{x\left(x^2-yz+2010\right)}+\frac{y^2}{y\left(y^2-xz+2010\right)}+\frac{z^2}{z\left(z^2-xy+2010\right)}\)
\(\ge\frac{\left(x+y+z\right)^3}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}\left(1\right)\)
Lưu ý nhé : \(x\left(x^2-yz+2010\right)=x\left(x^2+xy+zx+1340\right)>0\)
\(y\left(y^2-xz+2010\right)>0\)
\(z\left(z^2-xy+2010\right)>0\)
Ta có \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+xz\right)\right]\)
do dó \(x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)\) \(\)
=\(\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+zx\right)+2010\right]\)
=\(\left(x+y+z\right)^3\left(2\right)\)
Từ (1) zà (2) suy ra
vt \(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)
dấu = xảy ra khi zà chỉ khi \(x=y=z=\frac{\sqrt{2010}}{3}\)
Đặt \(a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\Rightarrow\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=1\end{matrix}\right.\)
\(K=\frac{\frac{1}{a}}{\sqrt{\frac{1}{bc}\left(1+\frac{1}{a^2}\right)}}+\frac{\frac{1}{b}}{\sqrt{\frac{1}{ac}\left(1+\frac{1}{b^2}\right)}}+\frac{\frac{1}{c}}{\sqrt{\frac{1}{ab}\left(1+\frac{1}{c^2}\right)}}\) \(=\frac{\frac{1}{a}}{\sqrt{\frac{a^2+1}{a^2bc}}}+\frac{\frac{1}{b}}{\sqrt{\frac{b^2+1}{ab^2c}}}+\frac{\frac{1}{c}}{\sqrt{\frac{c^2+1}{abc^2}}}\)
\(=\sqrt{\frac{bc}{a^2+1}}+\sqrt{\frac{ca}{b^2+1}}+\sqrt{\frac{ab}{c^2+1}}\) \(=\sqrt{\frac{bc}{a^2+ab+bc+ca}}+\sqrt{\frac{ca}{b^2+ab+bc+ca}}+\sqrt{\frac{ab}{c^2+ab+bc+ca}}\)
\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}+\frac{a}{a+b}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{b}{b+c}\right)\) \(\Rightarrow K\le\frac{3}{2}\)
Dấu "=" \(\Leftrightarrow a=b=c\Leftrightarrow x=y=z=\sqrt{3}\)
Áp dụng BĐT Cosi cho 2 sô dương ta có: \(x^2+yz\ge2x\sqrt{yz}\)
Tương tự: \(y^2+zx\ge2y\sqrt{zx};z^2+xy\ge2z\sqrt{xy}\)
Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được:
\(\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{zx}}+\frac{1}{2z\sqrt{xy}}\le\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\Leftrightarrow\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\le\frac{x+y+z}{xyz}\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le x+y+z\)
\(\Leftrightarrow\frac{1}{2}\left(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\right)\ge0\)(luôn đúng)
Dấu "=" xảy ra khi \(x=y=z\)
Áp dụng BĐT Cosi cho 2 số dương ta có: \(x^2+yz\ge2\sqrt{x^2yz}=2x\sqrt{yz}\)
Tương tự: \(y^2+zx\ge2y\sqrt{zx},z^2+xy\ge2z\sqrt{xy}\)
Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được:
\(\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{zx}}+\frac{1}{2z\sqrt{xy}}\le\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\Leftrightarrow\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\le\frac{x+y+z}{xyz}\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le x+y+z\)
\(\Leftrightarrow\frac{1}{2}\left(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\right)\ge0\)(luôn đúng)
Vậy BĐT được chứng minh. Dấu "=" xảy ra khi \(x=y=z\)
Cách 2:
Ta chuẩn hóa xyz=1
BĐT viết lại là \(\frac{x}{x^3+1}+\frac{y}{y^3+1}+\frac{z}{z^3+1}\le\frac{1}{2}\left(x+y+z\right)\)
Ta sử dụng đánh giá
\(x-\frac{2x}{x^3+1}+\frac{3}{2}\ge\frac{9x^2}{2\left(x^2+x+1\right)}\)\(\Leftrightarrow\frac{\left(x-1\right)^2\left(2x^4+3x^2+7x+3\right)}{2\left(x^3+1\right)\left(x^2+x+1\right)}\ge0\)
Do vậy ta cần c/m \(\frac{x^2}{x^2+x+1}+\frac{y^2}{y^2+y+1}+\frac{z^2}{z^2+z+1}\ge1\)
ta có \(\left(x;y;z\right)\rightarrow\left(\frac{a^2}{bc};\frac{b^2}{ca};\frac{c^2}{ab}\right)\)
BĐT viết lại là \(\frac{a^4}{a^4+a^2bc+\left(bc\right)^2}+\frac{b^4}{b^4+b^2ca+\left(ca\right)^2}+\frac{c^4}{c^4+c^2ab+\left(ab\right)^2}\ge1\)
Theo bđt Cauchy-Schwarz ta có
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+abc\left(a+b+c\right)+\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2}\)
Theo bđt AM-GM ta có
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+2\left(ab\right)^2+2\left(bc\right)^2+2\left(ca\right)^2}=1\)
Dấu "=" xảy ra khi a=b=c=> x=y=z
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)