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\(\sqrt{a+b}.\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)}\)
\(=\sqrt{2+\frac{a}{b}+\frac{b}{a}}\ge\sqrt{2+2\sqrt{\frac{a}{b}.\frac{b}{a}}}=\sqrt{2+2}=2\)
Dấu bằng xảy ra khi a = b.
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\(S=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1+4^2\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(1\right)\)\(\left(bunhia\right)\)
\(tương-tự\Rightarrow\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\left(2\right)\)
\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}+b+\dfrac{4}{c}+c+\dfrac{4}{a}\right)\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[2\sqrt{16a.\dfrac{4}{a}}+2\sqrt{16b.\dfrac{4}{b}}+2\sqrt{16c.\dfrac{4}{c}}-15.\dfrac{3}{2}\right]\left(am-gm\right)\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)
\(\Rightarrow MinS=\dfrac{3\sqrt{17}}{2}\Leftrightarrow a=b=c=\dfrac{1}{2}\)
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\(T=\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)
\(=\dfrac{8\left(a+b+c\right)}{9\sqrt[3]{abc}}+\dfrac{a+b+c}{9\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)
\(\ge\dfrac{8.3\sqrt[3]{abc}}{9\sqrt[3]{abc}}+2\sqrt{\dfrac{a+b+c}{9\sqrt[3]{abc}}.\dfrac{\sqrt[3]{abc}}{a+b+c}}=\dfrac{8}{3}+\dfrac{2}{3}=\dfrac{10}{3}\)
\(minT=\dfrac{10}{3}\Leftrightarrow a=b=c\)
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Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1^2+4^2\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)
\(\Leftrightarrow17\cdot\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)
\(\Leftrightarrow\sqrt{17}\cdot\sqrt{a^2+\frac{1}{b^2}}\ge a+\frac{4}{b}\)
Tương tự ta có :
\(\sqrt{17}\cdot\sqrt{b^2+\frac{1}{c^2}}\ge b+\frac{4}{c}\)
\(\sqrt{17}\cdot\sqrt{c^2+\frac{1}{a^2}}\ge c+\frac{4}{a}\)
Cộng theo vế của 3 bđt ta được :
\(\sqrt{17}\cdot\left(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\right)\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(\Leftrightarrow\sqrt{17}\cdot A\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
Áp dụng bất đẳng thức Cô-si :
\(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(=16a+\frac{4}{a}+16b+\frac{4}{b}+16c+\frac{4}{c}-15a-15b-15c\)
\(\ge2\sqrt{\frac{4\cdot16a}{a}}+2\sqrt{\frac{4\cdot16b}{b}}+2\sqrt{\frac{4\cdot16c}{c}}-15\left(a+b+c\right)\)
\(\ge16+16+16-15\cdot\frac{3}{2}=\frac{51}{2}\)
Do đó : \(\sqrt{17}\cdot A\ge\frac{51}{2}\)
\(\Leftrightarrow A\ge\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\)
@Nguyễn Việt Lâm
@Lê Thị Thục Hiền