Question

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Câu 1: Cho \(x,y,z>0\)thỏa mãn \(xyz=1\)

Chứng minh rằng : \(\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}\le\frac{1}{2}\)

Câu 2: Cho \(a,b,c>0\). Tìm min \(P=\frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}\)

Cây 3: Cho \(a,b,c>-1\). Chứng minh rằng :

\(\frac{1+a^2}{1+b+c^2}+\frac{1+b^2}{1+c+a^2}+\frac{1+c^2}{1+a+b^2}\ge2\)

Answer 1

Câu 1:

Áp dụng BĐT Cauchy ta có:

\(P=\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}=\frac{1}{(x^2+y^2)+(y^2+1)+2}+\frac{1}{(y^2+z^2)+(z^2+1)+2}+\frac{1}{(z^2+x^2)+(x^2+1)+2}\)

\(\leq \frac{1}{2xy+2y+2}+\frac{1}{2yz+2z+2}+\frac{1}{2zx+2x+2}\)

hay \(P\leq \frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)(1)\)

Do $xyz=1$ nên:

\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{1}{xy+y+1}+\frac{xy}{xy.yz+xyz+xy}+\frac{y}{yzx+yx+y}\)

\(=\frac{1}{xy+y+1}+\frac{xy}{y+1+xy}+\frac{y}{1+yx+y}=\frac{1+xy+y}{1+xy+y}=1(2)\)

Từ \((1);(2)\Rightarrow P\leq \frac{1}{2}.1=\frac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi $x=y=z=1$

Answer 2

Câu 2:

Đặt \((a+2b+c,a+b+2c,a+b+3c)=(x,y,z)\)

\(\Rightarrow \left\{\begin{matrix} c=-y+z\\ b=x-2y+z\\ a=-x+5y-3z\end{matrix}\right.\)

Khi đó, áp dụng BĐT Cauchy ta có:

\(P=\frac{-x+5y-3z-3y+3z}{x}+\frac{4x-8y+4z}{y}-\frac{-8y+8z}{z}\)

\(=-17+\left(\frac{2y}{x}+\frac{4x}{y}\right)+\left(\frac{4z}{y}+\frac{8y}{z}\right)\)

\(\geq -17+2\sqrt{\frac{2y}{x}.\frac{4x}{y}}+2\sqrt{\frac{4z}{y}.\frac{8y}{z}}=-17+12\sqrt{2}\)

Vậy \(P_{\min}=-17+12\sqrt{2}\)

Answer 3

Câu 3:

Áp dụng BĐT Cauchy:

\(b\leq |b|\leq \frac{b^2+1}{2}\Rightarrow \frac{a^2+1}{1+b+c^2}\geq \frac{a^2+1}{c^2+1+\frac{b^2+1}{2}}\). Tương tự với các phân thức còn lại:

\(\text{VT}=\frac{a^2+1}{1+b+c^2}+\frac{b^2+1}{1+c+a^2}+\frac{c^2+1}{1+a+b^2}\geq \frac{a^2+1}{\frac{b^2+1}{2}+c^2+1}+\frac{b^2+1}{\frac{c^2+1}{2}+a^2+1}+\frac{c^2+1}{\frac{a^2+1}{2}+b^2+1}\)

Đặt \((a^2+1,b^2+1,c^2+1)=(x,y,z)(x,y,z>0)\)

\(\text{VT}\geq \frac{x}{\frac{y}{2}+z}+\frac{y}{\frac{z}{2}+x}+\frac{z}{\frac{x}{2}+y}=2\left(\frac{x}{y+2z}+\frac{y}{z+2x}+\frac{z}{x+2y}\right)(1)\)

Cauchy-Schwarz:

\(\frac{x}{y+2z}+\frac{y}{z+2x}+\frac{z}{x+2y}=\frac{x^2}{xy+2xz}+\frac{y^2}{yz+2xy}+\frac{z^2}{xz+2yz}\geq \frac{(x+y+z)^2}{3(xy+yz+xz)}\geq \frac{(x+y+z)^2}{(x+y+z)^2}=1(2)\)

Từ \((1);(2)\Rightarrow \text{VT}\geq 2\) (đpcm)