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Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) khi đó thu được \(xyz=1\)
Ta có:
\(\dfrac{1}{a^2\left(b+c\right)}=\dfrac{x^2}{\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{x^2yz}{y+z}=\dfrac{x}{y+z}\)
BĐT cần chứng minh được viết lại thành:\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{y}{z+x}+1\right)+\left(\dfrac{z}{x+y}+1\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)\ge\dfrac{9}{2}\)
Đánh giá cuối cùng đúng theo BĐT Cauchy
Vậy BĐT được chứng minh. Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.
\(\dfrac{1}{a^3}+a\ge2\sqrt{\dfrac{a}{a^3}}=\dfrac{2}{a}\) ; \(\dfrac{1}{b^3}+b\ge\dfrac{2}{b}\) ; \(\dfrac{1}{c^3}+c\ge\dfrac{2}{c}\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (1)
Lại có \(\dfrac{4a}{a^4+1}\le\dfrac{4a}{2\sqrt{a^4}}=\dfrac{4a}{2a^2}=\dfrac{2}{a}\)
Tương tự \(\dfrac{4b}{b^4+1}\le\dfrac{2}{b}\) ; \(\dfrac{4c}{c^4+1}\le\dfrac{2}{c}\)
\(\Rightarrow4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\le2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (2)
Từ (1),(2)\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)
Dấu "=" xảy ra khi a=b=c=1
Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:
\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)
\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT AM - GM, ta có:
\(a^2+2b^2+3\)
\(=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)
\(\ge2ab+2b+2\)
Tương tự, ta có: \(b^2+2c^2+3\ge2bc+2c+2\) và \(c^2+2a^2+3\ge2ac+2a+2\)
\(VT=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)
\(\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a^2bc+abc}\right)\) (Thay abc = 1)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1+ab+b}{ab+b+1}\)
\(=\dfrac{1}{2}=VP\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c = 1
\(\sum\dfrac{a}{\left(a^2+1\right)+2b+2}\le\sum\dfrac{a}{2\left(a+b+1\right)}=\dfrac{1}{2}\)
Lời giải:
Áp dụng hệ quả của BĐT AM-GM:
\(\text{VT}^2=\left[\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\right]^2\geq 3\left(\frac{1}{ab(a+1)(b+1)}+\frac{1}{bc(b+1)(c+1)}+\frac{1}{ca(a+1)(c+1)}\right)\)
\(\Leftrightarrow \text{VT}^2\geq 3.\frac{a^2+b^2+c^2+a+b+c}{abc(a+1)(b+1)(c+1)}\geq 3.\frac{a+b+c+ab+bc+ac}{abc(a+1)(b+1)(c+1)}\)
\(\Leftrightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(a+1)(b+1)(c+1)}\) \((1)\)
Ta sẽ cm \((a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3\). Thật vậy:
Áp dụng BĐT AM-GM:
\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}\)
Cộng theo vế: \(\Rightarrow 3\geq \frac{3(\sqrt[3]{abc}+1)}{\sqrt[3]{(a+1)(b+1)(c+1)}}\)
\(\Rightarrow (a+1)(b+1)(c+1)\geq (\sqrt[3]{abc}+1)^3\) (2)
Từ \((1),(2)\Rightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(1+\sqrt[3]{abc})^3}=\frac{9}{\sqrt[3]{a^2b^2c^2}(1+\sqrt[3]{abc})^2}=\text{VP}^2\)
\(\Leftrightarrow \text{VT}\geq \text{VP}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
Ta có BĐT:
\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\)\(\ge\left(a+b\right)ab\)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)
\(\Rightarrow\dfrac{1}{a^3+b^3+abc}\le\dfrac{1}{ab\left(a+b+c\right)}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{ab\left(a+b+c\right)}+\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ac\left(a+b+c\right)}\)
\(=\dfrac{a+b+c}{abc\left(a+b+c\right)}=\dfrac{1}{abc}=VP\)
Khi \(a=b=c\)