Đặt \(A=x^2+y^2+z^2+xy+yz+zx\)

Áp dụng BĐT Bunyakovsky dạng phân thức, ta được: \(2A=x^2+y^2+z^2+\left(x+y+z\right)^2\ge\frac{\left(x+y+z\right)^2}{3}+\left(x+y+z\right)^2\)

\(=\frac{4\left(x+y+z\right)^2}{3}=12\Rightarrow A\ge6\)

Đẳng thức xảy ra khi x = y = z = 1

Ta có $xy+yz+zx=xyz$

$\Leftrightarrow\dfrac1x+\dfrac1y+\dfrac1z=1.$

Đặt $a=\dfrac1x,\quad b=\dfrac1y,\quad c=\dfrac1z.$

Khi đó $a+b+c=1.$

Ta có $H=\dfrac{c}{1+9a^2}+\dfrac{a}{1+9b^2}+\dfrac{b}{1+9c^2}.$

Với $t>0$, ta có $\dfrac1{1+9t^2}\ge1-\dfrac32t$ vì $\dfrac1{1+9t^2}-\left(1-\dfrac32t\right)$

$=\dfrac{\frac32t(3t-1)^2}{1+9t^2}\ge0.$

Do đó $H\ge c\left(1-\dfrac32a\right)+a\left(1-\dfrac32b\right)+b\left(1-\dfrac32c\right)$

$=a+b+c-\dfrac32(ab+bc+ca)$

$=1-\dfrac32(ab+bc+ca).$

Mặt khác, $(a+b+c)^2\ge3(ab+bc+ca)$

$\Rightarrow ab+bc+ca\le\dfrac13.$

Suy ra $H\ge1-\dfrac32\cdot\dfrac13$$=\dfrac12.$

Dấu bằng xảy ra khi $a=b=c=\dfrac13$

$\Leftrightarrow x=y=z=3.$

Vậy $H_{\min}=\dfrac12.$

\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)

Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)

\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)

Cộng vế:

\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)

Áp dụng BĐT Cô-si cho 2 số thực dương \(\dfrac{xy}{z}\) và \(\dfrac{yz}{x}\) có:

\(\dfrac{xy}{z}+\dfrac{yz}{x}\) \(\ge\) 2\(\sqrt{\dfrac{xy}{z}\cdot\dfrac{yz}{x}}\) = 2\(\sqrt{y^2}\) = 2y (1)

Tương tự: \(\dfrac{yz}{x}+\dfrac{zx}{y}\ge2z\) (2)

\(\dfrac{xy}{z}+\dfrac{zx}{y}\ge2x\) (3)

Từ (1); (2); (3)

\(\Rightarrow\) \(\dfrac{2xy}{z}+\dfrac{2yz}{x}+\dfrac{2zx}{y}\ge2x+2y+2z\)

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

\(\Leftrightarrow\) \(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\ge x+y+z=10\)

Hay P_Min = 10 

Dấu "=" xảy ra \(\Leftrightarrow\) x = y = z = \(\dfrac{10}{3}\)

Vậy ...

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