Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^4+b^4)(a^2+b^2)\geq (a^3+b^3)^2\)

\(\Rightarrow \frac{a^4+b^4}{ab(a^3+b^3)}\geq \frac{(a^3+b^3)^2}{ab(a^3+b^3)(a^2+b^2)}=\frac{a^3+b^3}{ab(a^2+b^2)}(1)\)

Tiếp tục áp dụng BĐT Bunhiacopxky:

\((a^3+b^3)(a+b)\geq (a^2+b^2)^2\)

Mà theo hệ quả BĐT AM-GM: \(a^2+b^2\geq \frac{(a+b)^2}{2}\)

Suy ra \((a^3+b^3)(a+b)\geq (a^2+b^2)\frac{(a+b)^2}{2}\)

\(\Leftrightarrow a^3+b^3\geq \frac{(a+b)(a^2+b^2)}{2}(2)\)

Từ (1); (2) suy ra \(\frac{a^4+b^4}{ab(a^3+b^3)}\geq \frac{a^3+b^3}{ab(a^2+b^2)}\geq \frac{a+b}{2ab}\)

Tương tự với các phân thức còn lại và cộng theo vế thu được:

\(\sum \frac{a^4+b^4}{ab(a^3+b^3)}\geq \frac{a+b}{2ab}+\frac{b+c}{2bc}+\frac{a+c}{2ac}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

Lời giải:

Ta có:

\(\sum \frac{1}{a+ab}\geq \frac{3}{abc+1}\Leftrightarrow \sum \frac{abc+1}{a(b+1)}\geq 3\)

\(\Leftrightarrow \sum \frac{bc}{b+1}+\sum\frac{1}{a(b+1)}\geq 3\)

\(\Leftrightarrow \sum \frac{b(c+1)}{b+1}+\sum \frac{a+1}{a(b+1)}\geq 6\)

BĐT trên luôn đúng vì theo BĐT AM-GM thì:

\(\sum \frac{b(c+1)}{b+1}+\sum \frac{a+1}{a(b+1)}=\frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1}+\frac{a+1}{a(b+1)}+\frac{b+1}{b(c+1)}+\frac{c+1}{c(a+1)}\)

\(\geq 6\sqrt[6]{\frac{abc(a+1)^2(b+1)^2(c+1)^2}{abc(a+1)^2(b+1)^2(c+1)^2}}=6\)

Do đó ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c=1\)

a.

Bình phương 2 vế, BĐT cần chứng minh trở thành:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)

Ta có:

\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)

Tương tự cộng lại:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b.

\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)

Nên ta chỉ cần chứng minh:

\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)

\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)

Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)

thay đề liên tục nhỉ

\(\sum\dfrac{1}{a}=\dfrac{\left(b^2+c^2\right)}{a\left(b^2+c^2\right)}+\dfrac{a^2+c^2}{b\left(a^2+c^2\right)}+\dfrac{b^2+a^2}{c\left(b^2+a^2\right)}\)

\(=\dfrac{b^2}{a\left(b^2+c^2\right)}+\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)

=\(\sum\left(\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\right)\ge\sum\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\) cauchy shawrtz

\(=\sum\dfrac{\left(a+b\right)^2}{a^2b+bc^2+ab^2+ac^2}=\sum\dfrac{\left(a+b\right)^2}{\left(a+b\right)\left(ab+c^2\right)}\)

\(=\sum\dfrac{a+b}{ab+c^2}\)(Q.E.D)

@Vũ Tiền Châu @Akai Haruma @Mysterious Person @Phùng Khánh Linh

Lời giải:

Áp dụng Cauchy-Schwarz kết hợp AM-GM
\(\frac{a+b}{bc+a^2}=\frac{(a+b)(b+c)}{(bc+a^2)(b+c)}=\frac{(a+b)(b+c)}{b(a^2+c^2)+c(a^2+b^2)}\)

\(\leq \frac{1}{2}\frac{(a+b)^2+(b+c)^2}{b(a^2+c^2)+c(a^2+b^2)}=\frac{1}{2}\left(\frac{(a+b)^2}{b(a^2+c^2)+c(a^2+b^2)}+\frac{(b+c)^2}{b(a^2+c^2)+c(a^2+b^2)}\right)\)

\(\leq \frac{1}{2}\left(\frac{a^2}{b(a^2+c^2)}+\frac{b^2}{c(a^2+b^2)}+\frac{b^2}{c(a^2+b^2)}+\frac{c^2}{b(a^2+c^2)}\right)\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:

\(\sum \frac{a+b}{bc+a^2}\leq \frac{a^2+c^2}{b(a^2+c^2)}+\frac{b^2+a^2}{c(a^2+b^2)}+\frac{c^2+b^2}{a(b^2+c^2)}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

@Akai Haruma @Vũ Tiền Châu @Phùng Khánh Linh

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:

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

\(\left\{{}\begin{matrix}a^2=x\\b^2=y\\c^2=z\end{matrix}\right.\Rightarrow VT=\dfrac{x^3}{y+z}+\dfrac{y^3}{x+z}+\dfrac{z^3}{x+y}=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{xy+yz}+\dfrac{z^4}{xz+yz}\)

\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{xy+yz+zx}{2}\)

\(xy+yz+zx=a^2b^2+b^2c^2+c^2a^2\)

Ta can cm \(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)

That vay

\(a^2b^2+b^2c^2\ge2b^2ac\)

\(b^2c^2+c^2a^2\ge2c^2ab\)

\(a^2b^2+a^2c^2\ge2a^2bc\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\left(dpcm\right)\)

\(\Rightarrow\sum\dfrac{a^6}{b^2+c^2}\ge\dfrac{abc\left(a+b+c\right)}{2}\)