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Lời giải:
Áp dụng BĐT Schur bậc 3:
\(abc\geq (a+b-c)(b+c-a)(c+a-b)\)
\(\Leftrightarrow abc\geq (1-2c)(1-2a)(1-2b)\)
\(\Leftrightarrow 9abc\geq 4(ab+bc+ac)-1\) (thay \(a+b+c=1\) )
\(\Rightarrow abc\geq \frac{4}{9}(ab+bc+ac)-\frac{1}{9}\)
\(\Rightarrow ab+bc+ac-abc\leq \frac{5}{9}(ab+bc+ac)+\frac{1}{9}\)
Theo hệ quả quen thuộc của BĐT AM-GM
\(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{1}{3}\Rightarrow \frac{5}{9}(ab+bc+ac)+\frac{1}{9}\leq \frac{8}{27}\)
\(\Rightarrow ab+bc+ac-abc\leq \frac{8}{27}\) (đpcm)
Dấu bằng xảy ra khi $3a=3b=3c=1$
À rồi, vừa mới nghĩ ra cách khác:
Ta có:
\(ab+bc+ac-abc=b(a+c)+ac-ac(1-a-c)\)
\(=b(a+c)+ac(a+c)=(a+c)(b+ac)=(a+c)(1-a-c+ac)=(a+c)(1-a)(1-c)\)
Áp dụng BĐT AM-GM:
\((a+c)(1-a)(1-c)\leq \left(\frac{a+c+1-a+1-c}{3}\right)^3=\frac{8}{27}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Có BĐT: \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Ta có:
\(VT=\)\(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)
\(=\dfrac{1+1+c^2}{\left(a^2+b^2+1\right)\left(1+1+c^2\right)}+\dfrac{1+1+a^2}{\left(b^2+c^2+1\right)\left(1+1+a^2\right)}+\dfrac{1+1+b^2}{\left(c^2+a^2+1\right)\left(1+1+b^2\right)}\)
Áp dụng BĐT Bunhiacopski cho mẫu số, ta có:
\(\left(a^2+b^2+c^2\right)\left(1+1+c^2\right)\ge\left(a+b+c\right)^2\)
\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(b+c+a\right)^2\)
\(\left(c^2+a^2+1\right)\left(1+1+b^2\right)\ge\left(c+a+b\right)^2\)
\(\Rightarrow VT\le\dfrac{1+1+c^2}{\left(a+b+c\right)^2}+\dfrac{1+1+a^2}{\left(b+c+a\right)^2}+\dfrac{1+1+b^2}{\left(c+a+b\right)^2}=\dfrac{6+a^2+b^2+c^2}{\left(a+b+c\right)^2}\le\dfrac{6+ab+bc+ca}{3\left(ab+bc+ca\right)}=\dfrac{6+3}{3.3}=1\)
\("="\Leftrightarrow a=b=c=1\)
\(\left\{{}\begin{matrix}\dfrac{1}{a+2}=\dfrac{1}{2}-\dfrac{1}{b+2}+\dfrac{1}{2}-\dfrac{1}{c+2}=\dfrac{b}{2\left(b+2\right)}+\dfrac{c}{2\left(c+2\right)}\ge\sqrt{\dfrac{bc}{\left(b+2\right)\left(c+2\right)}}\\\dfrac{1}{b+2}\ge\sqrt{\dfrac{ca}{\left(c+2\right)\left(a+2\right)}}\\\dfrac{1}{c+2}\ge\sqrt{\dfrac{ab}{\left(a+2\right)\left(b+2\right)}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\dfrac{abc}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(\Leftrightarrow abc\le1< \dfrac{9}{8}\)
Đề sai !
Giả sử \(a=b=c=1\) thay vào phương trình đầu thì :
\(\dfrac{1}{1+2}+\dfrac{1}{1+2}+\dfrac{1}{1+2}=1\) ( Thỏa mãn )
Nhưng \(1.1.1< \dfrac{1}{8}\) ( vô lí )
\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)