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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Cô si:
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+b}{8}\ge3\sqrt[3]{\frac{a^3}{\left(a+b\right)\left(b+c\right)}.\frac{\left(a+b\right)}{8}.\frac{\left(b+c\right)}{8}}=\frac{3a}{4}\)
Tương tự với 2 cục còn lại, công theo vế:
\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\frac{a+b+c}{4}\text{ }\left(dpcm\right)\)
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https://diendantoanhoc.net/topic/82335-cho-abc-la-d%E1%BB%99-dai-3-c%E1%BA%A1nh-c%E1%BB%A7a-tam-giac-co-chu-vi-b%E1%BA%B1ng-2-cmr-frac5227leq-a2b2c22abc-2/
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chuẩn hóa \(a^2+b^2+c^2=1\)
\(VT\ge\frac{3\sqrt{3}}{2}.\)
chúng ta cần chứng minh:\(\frac{a}{b^2+c^2}\ge\frac{3\sqrt{3}a^2}{2}\Leftrightarrow\frac{a}{1-a^2}\ge\frac{3\sqrt{3}a^2}{2}\)
\(\Leftrightarrow\frac{1}{1-a^2}\ge\frac{3\sqrt{3}a}{2}.\)
\(\Leftrightarrow a\left(1-a^2\right)\le\frac{2}{3\sqrt{3}}.\)
\(\Leftrightarrow a^2\left(1-a^2\right)^2\le\frac{4}{27}.\)
Mà\(\)
\(\Leftrightarrow2a^2\left(1-a^2\right)\left(1-a^2\right)\le\frac{\left(2a^2+1-a^2+1-a^2\right)^3}{27}=\frac{8}{27}.\left(dung\right)\)
Nên\(a^2\left(1-a^2\right)^2\le\frac{4}{27}\left(luondung\right)\)
Tương tự ta có: \(\frac{b}{a^2+c^2}\ge\frac{3\sqrt{3}b^2}{2};\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}c^2}{2}\)
Cộng lại ta có \(đpcm\)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
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