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\(3,\)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 bdt cauchy -schửat dạng engel ta có
\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)\(\ge\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2}=\frac{1}{2}\)
(do \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\) bn tự cm nhé)
dau = xay ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
Bạn kiểm tra lại đề
\(z=max\left\{x;y;z\right\}\)hay \(z=min\left\{x;y;z\right\}\)
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
Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
\(Q=\Sigma\frac{x^4}{x^2+\sqrt{xy.zx}}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+xy+yz+zx}\ge\frac{x^2+y^2+z^2}{2}\ge\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi x=y=z=1
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{xy}{\sqrt{z+xy}}=\frac{xy}{\sqrt{z\left(x+y+z\right)+xy}}=\frac{xy}{\sqrt{xz+yz+z^2+xy}}\)
\(=\frac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\frac{1}{2}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\frac{yz}{\sqrt{x+yz}}\le\frac{1}{2}\left(\frac{yz}{x+y}+\frac{yz}{x+z}\right);\frac{xz}{\sqrt{y+xz}}\le\frac{1}{2}\left(\frac{xz}{y+z}+\frac{xz}{x+y}\right)\)
Cộng theo vế các BĐT trên ta có:
\(P\le\frac{1}{2}\left(\frac{xy+yz}{x+z}+\frac{yz+xz}{x+y}+\frac{xy+xz}{y+z}\right)\)
\(=\frac{1}{2}\left(\frac{y\left(x+z\right)}{x+z}+\frac{z\left(x+y\right)}{x+y}+\frac{x\left(y+z\right)}{y+z}\right)\)
\(=\frac{1}{2}\left(x+y+z\right)=\frac{1}{2}\left(x+y+z=1\right)\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)