Chứng minh căn bậc 4(a/(b+c))+căn bậc 4(b/(c+a))+...>= căn bậc 4(...)

\(\sqrt[4]{\dfrac{a}{b+c}}+\sqrt[4]{\dfrac{b}{c+a}}+\sqrt[4]{\dfrac{c}{a+b}}\ge\sqrt[4]{16+\dfrac{196abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Leftrightarrow\sqrt[4]{\dfrac{a}{b+c}}+\sqrt[4]{\dfrac{b}{c+a}}+\sqrt[4]{\dfrac{c}{a+b}}\ge\sqrt[4]{\dfrac{16\left(a+b\right)\left(b+c\right)\left(c+a\right)+196abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Leftrightarrow\left(Σ\sqrt[4]{a\left(a+b\right)\left(a+c\right)}\right)^4\ge16\prod\left(a+b\right)+196\prod a\)

\(VT=Σa\left(a+b\right)\left(a+c\right)+4\left(Σ\sqrt[4]{\left(a\left(a+b\right)\left(a+c\right)\right)^3\left(b\left(b+c\right)\left(a+b\right)\right)}\right)\)

\(+6\left(Σ\sqrt[4]{\left(a\left(a+b\right)\left(a+c\right)\right)^2\left(b\left(b+c\right)\left(a+b\right)\right)^2}\right)\)

\(+4\left(\sqrt[4]{a\left(a+b\right)\left(a+c\right)\left(b\left(b+c\right)\left(a+b\right)\right)^3}\right)\)

\(+12Σ\sqrt[4]{\left(a\left(a+b\right)\left(a+c\right)\right)^2b\left(b+c\right)\left(a+b\right)c\left(c+a\right)\left(b+c\right)}\)

\(=\sum a(a+b)(a+c)+4\sum\sqrt[4]{(a^2(a+b+c)+abc)^3(b^2(a+b+c)+abc)}+\)

\(+4\sum\sqrt[4]{(a^2(a+b+c)+abc)^3(c^2(a+b+c)+abc)}\)

\(+6\sum\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\)

\(+12\sum\sqrt[4]{a^2bc(a+b)^3(a+c)^3(b+c)^2}\)

\(\ge\sum(a^3+a^2b+a^2c+abc)+4\sum\left(\left(\sqrt{a^3b}+\sqrt{a^3c}\right)(a+b+c)+2abc\right)\)

\(+6\sum(ab(a+b+c)+abc)+144abc\)

\(\ge\sum\left(a^3+7a^2b+7a^2c+4\sqrt{a^5b}+4\sqrt{a^5c}+8\sqrt{a^3b^3}+77abc\right)\)

\(\ge\sum\left(8a^2b+8a^2c+4\sqrt{a^5b}+4\sqrt{a^5c}+8\sqrt{a^3b^3}+76abc\right)\)

Vi` \(16\prod(a+b)+196abc=\sum(16^2b+16a^2c+76abc)\ge0\)

Ta can chung minh

\(\sum\left(4\sqrt{a^5b}+4\sqrt{a^5c}-8a^2b-8a^2c+8\sqrt{a^3b^3}\right)\ge0\)

\(\Leftrightarrow\sum\sqrt{ab}(a+b)(\sqrt{a}-\sqrt{b})^2\ge0\)