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Áp dụng BĐT AM-GM cho 3 số không âm, ta có: \(0< \sqrt[3]{yz.1}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{3x}{y+z+1}\)
Làm tương tự với 2 hạng tử còn lại rồi cộng theo vế thì có:
\(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{z+x+1}+\frac{z}{x+y+1}\right)\)
\(=3\left(\frac{x^2}{xy+xz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{zx+yz+z}\right)\ge^{Schwartz}3.\frac{\left(x+y+z\right)^2}{x+y+z+2\left(xy+yz+zx\right)}\)
\(=3.\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x+y+z+2\left(xy+yz+zx\right)}\ge9.\frac{xy+yz+zx}{\sqrt{3\left(x^2+y^2+z^2\right)}+2\left(x^2+y^2+z^2\right)}\)
\(=9.\frac{xy+yz+zx}{3+2.3}=xy+yz+zx\) => ĐPCM.
Dấu "=" xảy ra khi x=y=z=1.
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)
\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2}\)
\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Dấu "=" <=> x=y=z=1
\(\text{Σ}\sqrt{\frac{xy}{xy+z}}=\text{Σ}\sqrt{\frac{xy}{xy\left(x+y+z\right)}}=\text{Σ}\sqrt{\frac{xy}{\left(x+y\right)\left(x+z\right)}}\)
\(\le\text{Σ}\left(\frac{\frac{x}{x+y}+\frac{y}{x+z}}{2}\right)=\frac{3}{2}\)
Dấu = xảy ra khi x=y=z=1/3
a/ \(VT\ge\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{2\sqrt{a}}+\frac{\left(\sqrt{c}+\sqrt{a}\right)^2}{2\sqrt{b}}+\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\sqrt{c}}\)
\(VT\ge\frac{\left(\sqrt{b}+\sqrt{c}+\sqrt{c}+\sqrt{a}+\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ \(VT=\sum\frac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\)
\(VT\le\sum\frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Bài 1 :
Áp dụng BĐT Cô - si cho 2 số không âm ta có :
\(VT=\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\Sigma_{cyc}\sqrt{\frac{bc}{a}}\right)\)
\(\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)+\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)\)
\(\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(+3\sqrt[6]{abc}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt[3]{yz}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{x}{\frac{y+z+1}{3}}=\frac{3x}{y+z+1}\)
Tương tự rồi cộng lại ta có:
\(VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)\)
\(=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)\)
\(\ge\frac{3\left(x^4+y^4+z^4\right)}{2\left(xy+yz+xz\right)+x+y+z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}\)
\(=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Đẳng thức xảy ra khi \(x=y=z=1\)
Áp dụng BĐT AM-GM ta có:
\sqrt[3]{yz}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{x}{\frac{y+z+1}{3}}=\frac{3x}{y+z+1}3yz≤3y+z+1⇒3yzx≥3y+z+1x=y+z+13x
Tương tự rồi cộng lại ta có:
VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)VT≥3(y+z+1x+x+z+1y+x+y+1z)
=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)=3(xy+yz+xx2+xy+yz+yy2+yz+xz+zz2)
\ge\frac{3\left(x^4+y^4+z^4\right)}{2\left(xy+yz+xz\right)+x+y+z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}≥2(xy+yz+xz)+x+y+z3(x4+y4+z4)≥x2+y2+z2(x2+y2+z2)2
=x^2+y^2+z^2\ge xy+yz+xz=VP=x2+y2+z2≥xy+yz+xz=VP
Đẳng thức xảy ra khi x=y=z=1x=y=z=1