Xét \(n^2+1=n^2+mn+np+pm=n\left(m+n\right)+p\left(m+n\right)=\left(m+n\right)\left(n+p\right)\)

Tương tự: \(m^2+1=\left(m+n\right)\left(m+p\right)\)

\(p^2+1=\left(p+m\right)\left(p+n\right)\)

\(\Rightarrow\dfrac{\left(n^2+1\right)\left(p^2+1\right)}{m^2+1}=\dfrac{\left(n+p\right)^2\left(m+n\right)\left(m+p\right)}{\left(m+n\right)\left(m+p\right)}\)

\(=\left(n+p\right)^2\)

\(\Rightarrow\sqrt{\dfrac{\left(n^2+1\right)\left(p^2+1\right)}{m^2+1}}=n+p\)

Tương tự: \(\sqrt{\dfrac{\left(p^2+1\right)\left(m^2+1\right)}{n^2+1}}=m+p\)

\(\sqrt{\dfrac{\left(m^2+1\right)\left(n^2+1\right)}{p^2+1}}=m+n\)

\(\Rightarrow B=m\left(n+p\right)+n\left(m+p\right)+p\left(m+n\right)\)

\(=2\left(mn+np+pm\right)=2\)

Vậy B=2

MÌNH CẢM ƠN TRƯỚC NHA.

\(\Leftrightarrow\dfrac{z-mn}{m+n}-p+\dfrac{z-np}{n+p}-m+\dfrac{z-pm}{p+m}-n=0\)

\(\Leftrightarrow\dfrac{z-\left(mn+mp+np\right)}{m+n}+\dfrac{z-\left(mn+mp+np\right)}{n+p}+\dfrac{z-\left(mn+mp+np\right)}{p+m}=0\)

\(\Leftrightarrow\left[z-\left(mn+mp+np\right)\right]\left(\dfrac{1}{m+n}+\dfrac{1}{m+p}+\dfrac{1}{n+p}\right)=0\)

- Nếu \(\dfrac{1}{m+n}+\dfrac{1}{m+p}+\dfrac{1}{n+p}=0\) thì pt nghiệm đúng với mọi z

- Nếu \(\dfrac{1}{m+n}+\dfrac{1}{m+p}+\dfrac{1}{n+p}\ne0\)

\(\Rightarrow z=mn+mp+np\)

\(\dfrac{1}{a^3}+a\ge2\sqrt{\dfrac{a}{a^3}}=\dfrac{2}{a}\) ; \(\dfrac{1}{b^3}+b\ge\dfrac{2}{b}\) ; \(\dfrac{1}{c^3}+c\ge\dfrac{2}{c}\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (1)

Lại có \(\dfrac{4a}{a^4+1}\le\dfrac{4a}{2\sqrt{a^4}}=\dfrac{4a}{2a^2}=\dfrac{2}{a}\)

Tương tự \(\dfrac{4b}{b^4+1}\le\dfrac{2}{b}\) ; \(\dfrac{4c}{c^4+1}\le\dfrac{2}{c}\)

\(\Rightarrow4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\le2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (2)

Từ (1),(2)\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)

Dấu "=" xảy ra khi a=b=c=1