\(3,\)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

\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)

       \(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)

       \(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
        \(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)

\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)

Bài này phải tìm GTLN chứ nhỉ?!

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​⇒3yz​x​≥3y+z+1​x​=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

\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)

Ta có:\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\Rightarrow x+y+z=xyz\)

Dễ có một vài phép biến đổi cơ bản và bất đẳng thức AM - GM:\(\frac{x}{\sqrt{yz\left(1+x^2\right)}}=\frac{x}{\sqrt{yz+x^2yz}}=\frac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}\)

\(=\sqrt{\frac{x}{x+z}\cdot\frac{x}{x+y}}\le\frac{\frac{x}{x+z}+\frac{x}{x+y}}{2}\)

Khi đó:\(LHS\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{y}{x+y}+\frac{x}{x+z}+\frac{z}{x+z}+\frac{y}{z+y}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Đẳng thức xảy ra tại \(x=y=z=\sqrt{3}\)

\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)

\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)

\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)

\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

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 \(x=y=z=\frac{1}{3}\)