Cho $a,b,c>0$. Tìm giá trị nhỏ nhất của biểu thức: $P=\dfrac{a}{\sqrt{ab+b^2}}+\dfrac{b}{\sqrt{bc+c^2}}+\dfrac{c}{\sqrt{ca+a^2}}.$

Cân bằng hệ số dưới mẫu:

$\sqrt{ab+b^2}=\sqrt{b(a+b)}=\dfrac{1}{\sqrt{2}} \sqrt{2b(a+b)}$

Áp dụng BĐT Cauchy ta được:

$\sqrt{ab+b^2}\le \dfrac{1}{\sqrt{2}}.\sqrt{\left(\dfrac{2b+a+b}{2}\right)^2}=\dfrac{1}{\sqrt{2}}.\dfrac{3b+a}{2}=\dfrac{3b+a}{2\sqrt{2}}\\\to \dfrac{a}{\sqrt{ab+b^2}}\ge \dfrac{a}{\dfrac{3b+a}{2\sqrt{2}}}=\dfrac{2\sqrt{2}a}{a+3b}\\\to \sum\limits \dfrac{a}{\sqrt{ab+b^2}}\ge \sum\limits \dfrac{2\sqrt{2}a}{a+3b}=\sum\limits \dfrac{2\sqrt{2}a^2}{a^2+3ab}$

Thật vậy áp dụng BĐT Cauchy Schwars ta được:

$\sum\limits \dfrac{a^2}{a^2+3ab}\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+3(ab+bc+ac)}=\dfrac{(a+b+c)^2}{(a+b+c)^2+ab+bc+ac}$

Áp dụng BĐT Cauchy có:

$a^2+b^2\ge 2ab,b^2+c^2\ge 2bc,a^2+c^2\ge 2ac\\\to a^2+b^2+c^2\ge ab+bc+ac\\\to (a+b+c)^2\ge 3(ab+bc+ac)\\\to \dfrac{1}{3}(a+b+c)^2\ge ab+bc+ac\\\to\sum\limits \dfrac{a^2}{a^2+3ab}\ge \dfrac{(a+b+c)^2}{\dfrac{4}{3}(a+b+c)^2}=\dfrac{3}{4}\\\to \sum\limits \dfrac{a}{\sqrt{ab+b^2}}\ge 2\sqrt{2}.\dfrac{3}{4}=\dfrac{3\sqrt{2}}{2}$

Dấu "=" xảy ra khi: $a=b=c$

Ta có:

`P = a/\sqrt{ab + b^2} + b/\sqrt{bc + c^2} + c/\sqrt{ca + a^2}`

`= a^2/\sqrt{a^2(ab + b^2)} + b^2/\sqrt{b^2(bc + c^2)} + c^2/\sqrt{c^2(ca + a^2)}`

`= a^2/\sqrt{ab(a^2 + ab)} + b^2/\sqrt{bc(b^2 + bc)} + c^2/\sqrt{ac(c^2 + ac)}`

`= (\sqrt{2}a^2)/\sqrt{2ab(a^2 + ab)} + (\sqrt{2}b^2)/\sqrt{2bc(b^2 + bc)} + (\sqrt{2}c^2)/\sqrt{2ac(c^2 + ac)}`

Có: `\sqrt{2ab(a^2 + ab)} ≤ (2ab + a^2 + ab)/2 = (a^2 + 3ab)/2 \text{(BĐT Cô - si)}`

`⇒ (\sqrt{2}a^2)/\sqrt{2ab(a^2 + ab)} ≥` $\dfrac{\sqrt{2}a^2}{\dfrac{a^2 + 3ab}{2}} =$ `(2\sqrt{2}a^2)/(a^2 + 3ab)`

Chứng minh tương tự, ta được:

`(\sqrt{2}b^2)/\sqrt{2bc(b^2 + bc)} ≥ (2\sqrt{2}b^2)/(b^2 + 3bc)`

`(\sqrt{2}c^2)/\sqrt{2ac(c^2 + ac)} ≥ (2\sqrt{2}c^2)/(c^2 + 3ac)`

`⇒ P ≥ (2\sqrt{2}a^2)/(a^2 + 3ab) + (2\sqrt{2}b^2)/(b^2 + 3bc) + (2\sqrt{2}c^2)/(c^2 + 3ac) = 2\sqrt{2}(a^2/(a^2 + 3ab) + b^2/(b^2 + 3bc) + c^2/(c^2 + 3ac))`

Áp dụng BĐT $\text{Svacxơ}$, ta có:

`a^2/(a^2 + 3ab) + b^2/(b^2 + 3bc) + c^2/(c^2 + 3ac) ≥ (a + b + c)^2/(a^2 + b^2 + c^2 + 3ab + 3bc + 3ac) = (a + b + c)^2/((a + b + c)^2 + ab + ac + bc)`

Mà: `ab + bc + ac ≤ (a + b + c)^2/3`

`⇒ (a + b + c)^2 + ab + ac + bc ≤ (a + b + c)^2 + (a + b + c)^2/3 = [4(a + b + c)^2]/3`

`⇒ (a + b + c)^2/((a + b + c)^2 + ab + ac + bc) ≥` $\dfrac{(a + b + c)^2}{\dfrac{4(a + b + c)^2}{3}}$ `= 3/4`

`⇒ P ≥ 2\sqrt{2}(a^2/(a^2 + 3ab) + b^2/(b^2 + 3bc) + c^2/(c^2 + 3ac)) ≥ 2\sqrt{2} . 3/4 = (3\sqrt{2})/2`

Vậy GTNN của `P` là `(3\sqrt{2})/2` khi: `a = b = c`