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\(\dfrac{4}{3}=a+2\sqrt{\dfrac{a}{4}.b}+\dfrac{1}{2}\sqrt[3]{\dfrac{a}{2}.2b.8c}\)
\(\dfrac{4}{3}\le a+\dfrac{a}{4}+b+\dfrac{1}{6}\left(\dfrac{a}{2}+2b+8c\right)=\dfrac{4}{3}\left(a+b+c\right)\)
\(\Rightarrow a+b+c\ge1\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)
Anh ơi cho em hỏi làm sao để tách/tìm điểm rơi như thế này ạ?
\(P=\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge\dfrac{ab+bc+ca}{ab+bc+ca}=1\)
\(P_{min}=1\) khi \(a=b=c=1\)
\(P=\dfrac{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}{ab+bc+ca}=\dfrac{9}{ab+bc+ca}-2\)
Do \(a;b\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1=2-c\)
\(\Rightarrow ab+c\left(a+b\right)\ge2-c+c\left(3-c\right)=-c^2+2c+2=c\left(2-c\right)+2\ge2\)
\(\Rightarrow P\le\dfrac{9}{2}-2=\dfrac{5}{2}\)
\(P_{max}=\dfrac{5}{2}\) khi \(\left(a;b;c\right)=\left(1;2;0\right);\left(2;1;0\right)\)
Ta có:\(A\ge\left(a+b+1\right)\frac{\left(a+b\right)^2}{2}+\frac{4}{a+b}\)
Đặt \(t=a+b\)thì \(t\ge2\) theo AM-GM
Ta có:\(A\ge\frac{t^3}{2}+\frac{t^2}{2}+\frac{4}{t}=\frac{t^3}{2}+\frac{t^2}{4}+\frac{t^2}{4}+\frac{2}{t}+\frac{2}{t}\ge4+1+3=8\)
Đẳng thức xảy ra khi \(a=b=1\)
\(A=\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\ge\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)}+\frac{4}{ab\left(a+b\right)}\)
\(\ge\left(\frac{1}{a^2-ab+b^2}+\frac{1}{ab}+\frac{1}{ab}+\frac{1}{ab}\right)+\frac{1}{ab}\)
\(\ge\frac{\left(1+1+1+1\right)^2}{\left(a+b\right)^2}+\frac{1}{ab}\ge\frac{16}{\left(a+b\right)^2}+\frac{1}{\frac{\left(a+b\right)^2}{4}}\ge16+4=20\)
Đẳng thức xảy ra khi \(a=b=\frac{1}{2}\)
\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)
\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)
\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)
Dấu "=" xảy ra khi \(a=b=2\)
Ta có: \(1=4\left(a+b\right)+\sqrt{ab}\ge4.2\sqrt{ab}+\sqrt{ab}=9\sqrt{ab}\Leftrightarrow\sqrt{ab}\le\dfrac{1}{9}\Leftrightarrow ab\le\dfrac{1}{81}\)
\(\Rightarrow\dfrac{1}{ab}\ge\dfrac{1}{\dfrac{1}{81}}=81\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{1}{9}\)