Không mất tính tổng quát, giả sử \(x\ge y\ge z\)

\(y^2-yz+z^2=y^2+\left(z-y\right)y\le y^2\Rightarrow\dfrac{1}{y^2-yz+z^2}\ge\dfrac{1}{y^2}\)

Tương tự: \(\dfrac{1}{z^2-xz+x^2}\ge\dfrac{1}{x^2}\)

\(\Rightarrow P\ge\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{x^2-xy+y^2}+\dfrac{x^2-xy+y^2}{x^2y^2}+\dfrac{1}{xy}\)

\(P\ge2\sqrt{\dfrac{x^2-xy+y^2}{x^2y^2\left(x^2-xy+y^2\right)}}+\dfrac{1}{xy}=\dfrac{3}{xy}\ge\dfrac{12}{\left(x+y\right)^2}\ge\dfrac{12}{\left(x+y+z\right)^2}=3\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\) và hoán vị

Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)

Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)

\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)

Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)

\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)

\(\Rightarrow A\ge\dfrac{1}{2}\)

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)

Bài 1:

Áp dụng BĐT AM-GM cho các số thực dương ta có:

\(\frac{x^2}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^2}{4}}=x\)

\(\frac{y^2}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^2}{4}}=y\)

\(\frac{z^2}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^2}{4}}=z\)

Cộng theo vế:

\(\Rightarrow M+\frac{y+z}{4}+\frac{x+z}{4}+\frac{x+y}{4}\geq x+y+z\)

\(\Leftrightarrow M\geq \frac{x+y+z}{2}=\frac{2}{2}=1\)

Vậy GTNN của $M$ là $1$. Đẳng thức xảy ra tại $x=y=z=\frac{2}{3}$

Bài 2:

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

\(=(a+b+c+3)-\left(\frac{b^2(a+1)}{b^2+1}+\frac{c^2(b+1)}{c^2+1}+\frac{a^2(c+1)}{a^2+1}\right)\)

\(=6-M(*)\)

Xét \(M=\frac{b^2(a+1)}{b^2+1}+\frac{c^2(b+1)}{c^2+1}+\frac{a^2(c+1)}{a^2+1}\). Áp dụng BĐT AM-GM:

\(M\leq \frac{b^2(a+1)}{2b}+\frac{c^2(b+1)}{2c}+\frac{a^2(c+1)}{2a}=\frac{ab+bc+ac+a+b+c}{2}=\frac{ab+bc+ac+3}{2}\)

\(\leq \frac{\frac{(a+b+c)^2}{3}+3}{2}=3(**)\)

Từ \((*); (**)\Rightarrow \text{VT}=6-M\geq 6-3=3\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$

Cách khác bài 1:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\right)(y+z+x+z+x+y)\geq (x+y+z)^2\)

\(\Rightarrow M=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\geq \frac{(x+y+z)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}=\frac{2}{2}=1\)

Vậy GTNN của $M$ là $1$. Đẳng thức xảy ra khi $x=y=z=\frac{2}{3}$

\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2^2-2.1=2\) (đpcm)

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Lời giải:

Áp dụng BĐT Cô-si:

\(x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}\)

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{1}{3}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2\geq \frac{1}{3}.(\frac{9}{x+y+z})^2=\frac{27}{(x+y+z)^2}\)

\(\Rightarrow P\geq \frac{(x+y+z)^2}{3}+\frac{27}{(x+y+z)^2}\)

Áp dụng BĐT Cô-si:

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

\(\frac{80}{3(x+y+z)^2}\geq \frac{80}{3}\)

\(\Rightarrow P\geq \frac{2}{3}+\frac{80}{3}=\frac{82}{3}\)

Vậy $P_{\min}=\frac{82}{3}$ khi $x=y=z=\frac{1}{3}$