\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)

Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)

\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)

Cộng vế:

\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)

\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)

Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)

\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)

Cộng vế:

\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)

\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

chết đăng nhầm sogy nha

Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)

BĐT trở thành:

\(\dfrac{y^2}{xz}+\dfrac{z^2}{xy}+\dfrac{x^2}{yz}\ge\dfrac{3}{2}\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}-1\right)\)

\(\Leftrightarrow2\left(x^3+y^3+z^3\right)+3xyz\ge3x^2y+3y^2z+3z^2x\)

Áp dụng BĐT Schur ta có:

\(x^3+y^3+z^3+3xyz\ge x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\)

\(\Rightarrow VT\ge\left(x^3+xy^2\right)+\left(y^3+yz^2\right)+\left(z^3+zx^2\right)+x^2y+y^2z+z^2x\ge3\left(x^2y+y^2z+z^2x\right)\)

ta có \(\dfrac{1}{\left(a+b\right)c}\le\dfrac{1}{2\sqrt{ab}c}=\dfrac{1}{2\sqrt{c}}\)tương tự ta có

\(\Sigma\dfrac{1}{\left(a+b\right)c}\le\Sigma\dfrac{1}{2\sqrt{c}}=\dfrac{\Sigma\sqrt{ab}}{2}\le\dfrac{\Sigma a}{2}\)(đpcm)

Áp dụng BĐT Cauchy-Schwarz ta có:

\((ab+a+1)^2 \le (a+b+c) \left( a+ a^2b+ \frac 1c \right) = (a+b+c)(a+a^2b+ab)\)

\(\Rightarrow \dfrac{a}{(ab+a+1)^2} \ge \dfrac{a}{(a+b+c)(a+a^2b+ab)}= \dfrac{1}{(a+b+c)(1+ab+b)}\)

Thiết lập các BĐT tương tự rồi cộng theo vế ta có:

\(\sum \dfrac{a}{(ab+a+1)^2} \ge \dfrac{1}{a+b+c} \sum \dfrac{1}{ab+b+1}= \dfrac{1}{a+b+c}\)

c₂: Áp dụng BĐT bunyakovsky:

\(\left(a+b+c\right)\left[\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right]\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\right)^2\)

Xét \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)

\(=\dfrac{ab+a+1}{ab+a+1}=1\)

do đó \(\left(a+b+c\right).VT\ge1\Leftrightarrow VT\ge\dfrac{1}{a+b+c}\)

dấu = xảy ra khi a=b=c=1

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

\(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.

Nhìn qua đã biết là đề sai rồi bạn

Cho \(a,b,c\) các giá trị lớn ví dụ \(a=b=c=2\) là thấy sai ngay

Lời giải:

Áp dụng BĐT AM-GM ta có:

\(\frac{a^3}{(b+2)(c+3)}+\frac{b+2}{36}+\frac{c+3}{48}\geq 3\sqrt[3]{\frac{a^3}{36.48}}=\frac{a}{4}\)

Tương tự:\(\frac{b^3}{(c+2)(a+3)}+\frac{c+2}{36}+\frac{a+3}{48}\geq \frac{b}{4}\)

\(\frac{c^3}{(a+2)(b+3)}+\frac{a+2}{36}+\frac{b+3}{48}\geq \frac{c}{4}\)

Cộng theo vế các BĐT trên và rút gọn ta có:

\(\frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\)

Mà cũng theo AM-GM:

\(a+b+c\geq 3\sqrt[3]{abc}=3\)

\(\Rightarrow \frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\geq \frac{29}{144}.3-\frac{17}{48}=\frac{1}{4}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$