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a) Ta có BĐT:
\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)ab\)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b+c\right)}\)
Tương tự cho 2 bất đẳng thức còn lại rồi cộng theo vế:
\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)
\(=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}=VP\)
Khi \(a=b=c\)
Cosi + Svac-xơ
Có : \(3=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(a+b+c\le3\)
\(\frac{1}{4-\sqrt{ab}}+\frac{1}{4-\sqrt{bc}}+\frac{1}{4-\sqrt{ca}}\le\frac{1}{4-\frac{a+b}{2}}+\frac{1}{4-\frac{b+c}{2}}+\frac{1}{4-\frac{c+a}{2}}\)
\(=-\left(\frac{1}{\frac{a+b}{2}-4}+\frac{1}{\frac{b+c}{2}-4}+\frac{1}{\frac{c+a}{2}-4}\right)\le\frac{-\left(1+1+1\right)^2}{a+b+c-12}=\frac{-9}{3-12}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
Áp dụng bđt : x^2+y^2+z^2 >= (x+y+z)^2/3 ta có :
\(\frac{\sqrt{b^2+2a^2}}{ab}\)= \(\frac{\sqrt{a^2+b^2+a^2}}{ab}\)>= \(\frac{\sqrt{\frac{\left(a+b+a\right)^2}{3}}}{ab}\) = \(\frac{2a+b}{\sqrt{3}ab}\) = \(\frac{2}{\sqrt{3}b}+\frac{1}{\sqrt{3}a}\)
Tương tự : \(\frac{\sqrt{c^2+2b^2}}{bc}\)>= \(\frac{2}{\sqrt{3}c}+\frac{1}{\sqrt{3}b}\) ; \(\frac{\sqrt{a^2+2c^2}}{ac}\)>= \(\frac{2}{\sqrt{3}a}+\frac{1}{\sqrt{3}c}\)
=> \(\frac{\sqrt{b^2+2a^2}}{ab}\)+ \(\frac{\sqrt{c^2+2b^2}}{bc}\)+ \(\frac{\sqrt{a^2+2c^2}}{ac}\)>= \(\frac{3}{\sqrt{3}a}+\frac{3}{\sqrt{3}b}+\frac{3}{\sqrt{3}c}\)
= \(\frac{3}{\sqrt{3}}\).(1/a+1/b+1/c) = \(\sqrt{3}\).(ab+bc+ca)/abc = \(\sqrt{3}\).abc/abc = \(\sqrt{3}\)
Dấu "=" xảy ra <=> a=b=c=3
=> ĐPCM
k mk nha
\(VT=\sum\frac{ab}{\sqrt{\left(a+b+c\right)c+ab}}=\sum\frac{ab}{\sqrt{\left(b+c\right)\left(c+a\right)}}\le\sum\frac{ab}{2}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{1}{2}\left[\frac{ab+ca}{b+c}+\frac{ab+bc}{c+a}+\frac{bc+ca}{a+b}\right]=\frac{1}{2}\left(a+b+c\right)=1\)
Đây nhé @Liana
\(2a^2+\left(2-\sqrt{3}\right)b^2+2a^2+\left(2-\sqrt{3}\right)c^2+\left(\sqrt{3}-1\right)b^2+\left(\sqrt{3}-1\right)c^2\)
\(\ge2\sqrt{4-2\sqrt{3}}ab+2\sqrt{4-2\sqrt{3}}ac+2\left(\sqrt{3}-1\right)bc\)
\(\Leftrightarrow4a^2+b^2+c^2\ge2\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)\)
\(\Leftrightarrow4\ge2\left(\sqrt{3}-1\right)\left(ab+bc+ca\right)\)
\(\Leftrightarrow1+\sqrt{3}\ge ab+bc+ca\)
\(\Rightarrow dpcm\)