\(\dfrac{1}{a^3}+1+1\ge\dfrac{3}{a}\) ; \(\dfrac{a^3}{b^3}+1+1\ge\dfrac{3a}{b}\) ; \(b^3+1+1\ge3b\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{a^3}{b^3}+b^3+6\ge3\left(\dfrac{1}{a}+\dfrac{a}{b}+b\right)=\left(\dfrac{1}{a}+\dfrac{a}{b}+b\right)+2\left(\dfrac{1}{a}+\dfrac{a}{b}+b\right)\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{a^3}{b^3}+b^3+6\ge\dfrac{1}{a}+\dfrac{a}{b}+b+2.3\sqrt[3]{\dfrac{ab}{ab}}\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{a^3}{b^3}+b^3\ge\dfrac{1}{a}+\dfrac{a}{b}+b\)

Dấu "=" xảy ra khi \(a=b=1\)

Nice proof, nhưng đã quy đồng là phải thế này :v

\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)

\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)

\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)

\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)

Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:

\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)

Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)

Áp dụng BĐT này ta có:

\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)

Cộng theo vế 3 BĐT trên ta có:

\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)

áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)

tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)

\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)

suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

suy ra dpcm

dau = xay ra khi a=b=c

\(VT=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)

\(=\dfrac{1}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3>=\dfrac{9}{2}-3=\dfrac{3}{2}\)

Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:

\(VT=\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\ge\dfrac{3abc}{2abc}+\dfrac{2ab}{c^2+ab}+\dfrac{2bc}{a^2+bc}+\dfrac{2ac}{b^2+ac}=\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\)

Áp dụng bất đẳng thức \(Cauchy-Schwarz\) \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}=\dfrac{a^2b^2}{c^2ab+a^2b^2}+\dfrac{b^2c^2}{a^2bc+b^2c^2}+\dfrac{a^2c^2}{b^2ac+a^2c^2}\ge\dfrac{\left(ab+bc+ac\right)^2}{c^2ab+a^2b^2+a^2bc+b^2c^2+b^2ac+a^2c^2}\)

Đặt: \(\left\{{}\begin{matrix}ab=x\\bc=y\\ac=z\end{matrix}\right.\) ta được: \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+xy+xz+xy}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)

Nên: \(\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\ge\dfrac{3}{2}+2.\dfrac{3}{2}=\dfrac{9}{2}\)

Mà: \(VT\ge\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\Leftrightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Lời giải:

Áp dụng BĐT AM-GM ta có: \(\frac{a^3+b^3+c^3}{2abc}\geq \frac{3\sqrt[3]{a^3b^3c^3}}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\) (1)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq \frac{(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2}{a^2+b^2+c^2+ab+bc+ac}\) (2)

Có:

\((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2=2(a^2+b^2+c^2)+2\sqrt{(a^2+b^2)(b^2+c^2)}+2\sqrt{(b^2+c^2)(c^2+a^2)}+\sqrt{(a^2+b^2)(c^2+a^2)}\)

Áp dụng BĐT Bunhiacopxky:

\(\sqrt{(a^2+b^2)(b^2+c^2)}\geq \sqrt{(ac+b^2)^2}=ac+b^2\)

\(\sqrt{(b^2+c^2)(c^2+a^2)}\geq \sqrt{(ba+c^2)^2}=ba+c^2\)

\(\sqrt{(a^2+b^2)(c^2+a^2)}\geq \sqrt{(a^2+bc)^2}=a^2+bc\)

\(\Rightarrow (\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 2(a^2+b^2+c^2)+2(a^2+b^2+c^2+ab+bc+ac)\)

\(\geq a^2+b^2+c^2+ab+bc+ac+2(a^2+b^2+c^2+ab+bc+ac)\) (AM-GM)

Hay \((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 3(a^2+b^2+c^2+ab+bc+ac)\) (3)

Từ \((2); (3)\Rightarrow \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq 3\) (4)

Từ \((1); (4)\Rightarrow \frac{a^3+b^3+c^3}{2abc}+\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ac}\geq \frac{9}{2}\)

Ta có đpcm.

Dấu bằng xảy ra khi $a=b=c$

Câu a)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq \frac{9}{a+2b}\) (1)

\(\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\geq \frac{9}{b+2c}\)(2)

\(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\geq \frac{9}{c+2a}\) (3)

Lấy \((1)+2.(2)+3.(3)\) ta có:

\(\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{2}{b}+\frac{2}{c}+\frac{2}{c}+\frac{3}{c}+\frac{3}{a}+\frac{3}{a}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)

\(\Leftrightarrow \frac{7}{a}+\frac{4}{b}+\frac{7}{c}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

Câu b)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a}+\frac{4}{b}\geq \frac{(1+2)^2}{a+b}=\frac{9}{a+b}\)

\(\Rightarrow \frac{1}{3a}+\frac{4}{3b}\geq \frac{3}{a+b}(1)\)

\(\frac{1}{3b}+\frac{1}{2c}+\frac{1}{2c}\geq \frac{9}{3b+4c}\)

\(\Rightarrow \frac{2}{3b}+\frac{2}{c}\geq \frac{18}{3b+4c}\) (2)

\(\frac{1}{c}+\frac{1}{3a}+\frac{1}{3a}\geq \frac{9}{c+6a}\) (3)

Từ (1); (2); (3) cộng theo vế:

\(\Rightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{3}{a+b}+\frac{18}{3b+4c}+\frac{9}{c+6a}\)

(đpcm)

Dấu bằng xảy ra khi \(a=\frac{b}{2}=\frac{c}{3}\)

Câu c)

BĐT cần chứng minh tương đương với:
\(\frac{b+c+a}{a}+\frac{2a+c}{b}+\frac{4(a+b)}{a+c}\geq 10\) (*)

Áp dụng BĐT AM-GM:

\(\text{VT}=\frac{b}{a}+\frac{c+a}{2a}+\frac{c+a}{2a}+\frac{a}{b}+\frac{a+c}{2b}+\frac{a+c}{2b}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}\)

\(\geq 10\sqrt[10]{\frac{ba(c+a)^4(a+b)^4}{16a^3b^3(a+c)^4}}=10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\)

Theo AM-GM: \((a+b)^2\geq 4ab\Rightarrow (a+b)^4\geq 16a^2b^2\)

\(\Rightarrow \text{VT}\geq 10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\geq 10\)

Vậy (*) được cm. Ta có đpcm. Dấu bằng xảy ra khi a=b=c