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Áp dụng bđt Schwarz ta có:
\(P=\dfrac{a^4}{2ab+3ac}+\dfrac{b^4}{2cb+3ab}+\dfrac{c^4}{2ac+3bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{5\left(a^2+b^2+c^2\right)}=\dfrac{1}{5}\).
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\dfrac{\sqrt{3}}{3}\).
Theo BĐT Cauchy: \(\left\{{}\begin{matrix}\sqrt{ab}\le\dfrac{a}{4}+b\\\sqrt[3]{abc}\le\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)\end{matrix}\right.\)
\(\Rightarrow a+\sqrt{ab}+\sqrt[3]{abc}\le a+\dfrac{a}{4}+b+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)=\dfrac{4}{3}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1346}{\dfrac{4}{3}\left(a+b+c\right)}-\dfrac{2019}{\sqrt{a+b+c}}=\dfrac{2019}{2\left(a+b+c\right)}-\dfrac{2019}{\sqrt{a+b+c}}\)
\(\Rightarrow\dfrac{2P}{2019}\ge\dfrac{1}{a+b+c}-\dfrac{2}{\sqrt{a+b+c}}=\left(\dfrac{1}{\sqrt{a+b+c}}\right)^2-2.\dfrac{1}{\sqrt{a+b+c}}+1-1\)
\(\Rightarrow\dfrac{2P}{2019}\ge\left(\dfrac{1}{\sqrt{a+b+c}}-1\right)^2-1\ge-1\)
\(\Rightarrow P\ge\dfrac{-2019}{2}\)
\(\Rightarrow P_{min}=\dfrac{-2019}{2}\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\dfrac{a}{4}=b=4c\\\dfrac{1}{\sqrt{a+b+c}}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{16}{21}\\b=\dfrac{4}{21}\\c=\dfrac{1}{21}\end{matrix}\right.\)
\(VT=\sqrt{\dfrac{a^2b^2}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{b^2c^2}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{a^2c^2}{b\left(a+b+c\right)+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{ac+ab+bc+c^2}}+\sqrt{\dfrac{b^2c^2}{a^2+ac+ab+bc}}+\sqrt{\dfrac{a^2c^2}{ab+bc+b^2+ac}}\\ VT=\sqrt{\dfrac{a^2b^2}{\left(c+a\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2c^2}{\left(b+c\right)\left(a+b\right)}}+\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{bc}{a+b}+\dfrac{bc}{a+c}}{2}\\\sqrt{\dfrac{a^2c^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(b+c\right)\left(a+c\right)}}\le\dfrac{\dfrac{ab}{b+c}+\dfrac{ab}{a+c}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ca}{b+c}+\dfrac{ab}{b+c}\right)+\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)}{2}\\ \Rightarrow VT\le\dfrac{a+b+c}{2}=\dfrac{2}{2}=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
a: \(A=2^{\dfrac{1}{3}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{3}+\dfrac{2}{3}}=2^{\dfrac{3}{3}}=2^1=2\)
b: \(B=36^{\dfrac{3}{2}}=\left(6^2\right)^{\dfrac{3}{2}}=6^{2\cdot\dfrac{3}{2}}=6^3=216\)
c: \(C=36^{\dfrac{3}{2}}\cdot\left(\dfrac{1}{6}\right)^2=\left(6^2\right)^{\dfrac{3}{2}}\cdot\dfrac{1}{6^2}=\dfrac{6^{2\cdot\dfrac{3}{2}}}{6^2}=\dfrac{6^3}{6^2}=6\)
d: \(D=\sqrt{81}\cdot\left(\dfrac{1}{3}\right)^2=9\cdot\dfrac{1}{3^2}=9\cdot\dfrac{1}{9}=1\)
e: \(E=\left(3+2\sqrt{2}\right)^{50}\cdot\left(3-2\sqrt{2}\right)^{50}\)
\(=\left[\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)\right]^{50}\)
\(=\left(9-8\right)^{50}=1^{50}=1\)
f: \(F=120^{\sqrt{5}+1}\cdot120^{3-\sqrt{5}}\)
\(=120^{\sqrt{5}+1+3-\sqrt{5}}=120^4\)
g: \(G=\left(3+2\sqrt{2}\right)^{2019}\cdot\left(3\sqrt{2}-4\right)^{2018}\)
\(=\left(3+2\sqrt{2}\right)^{2018}\cdot\left(3\sqrt{2}-4\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)
\(=\left[\left(3+2\sqrt{2}\right)\left(3\sqrt{2}-4\right)\right]^{2018}\left(3+2\sqrt{2}\right)\)
\(=\left(9\sqrt{2}-12+12-8\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)
\(=\left(\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)=2^{\dfrac{1}{2}\cdot2018}\cdot\left(3+2\sqrt{2}\right)\)
\(=2^{1009}\cdot\left(3+2\sqrt{2}\right)\)
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