[Toán 9]Bất đẳng thức

[rua_it]

Let a,b,c are positive real numbers.Prove that:
[TEX]\sum \frac {a^3}{b^2 - bc + c^2} \ge \frac {3(ab + bc + ac)}{a + b + c}[/TEX]

Kéo topic này lên
Theo Cauchy-Schwarz, ta có: [tex](\sum a^2)^2=(\sum \sqrt{\frac{a^3}{b^2-bc+c^2}}.\sqrt{(b^2-bc+c^2).a})^2 [/tex]
[tex]\leq VT.\sum a.(b^2-bc+c^2) \Rightarrow VT \geq \frac{(a^2+b^2+c^2)^2}{a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)-3abc}[/tex]
Mặt khác, ta lại có: [tex]VP =\frac{3(ab+bc+ca)}{a+b+c} \leq \frac{(a+b+c)^2}{a+b+c}=a+b+c[/tex]
Do đó, ta chỉ cần chứng minh [tex]\frac{(a^2+b^2+c^2)^2}{a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)-3abc} \geq a+b+c(*)[/tex]
Thật vậy, giả sử [tex]a \geq b \geq c \geq0[/tex]
[tex]\Rightarrow (*) \Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2 \geq (a+b+c)(a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)-3abc)[/tex]
[tex]\Leftrightarrow a^2(a-b)^2+(a+b)(a-c)(a-b)^2+c^2(c-a)(c-b) \geq 0[/tex]
[tex]\Rightarrow (*) [/tex] đúng \Rightarrow dpcm

 

[6262127]

Cho [tex]a,b,c>0;ab+bc+ca=abc[/tex]
CMR [tex]\sum \frac{sqrt{b^2+2a^2}}{ab} \geq \sqrt{3}[/tex]

 

[bigbang195]

Cho [tex]a,b,c>0;ab+bc+ca=abc[/tex]
CMR [tex]\sum \frac{sqrt{b^2+2a^2}}{ab} \geq \sqrt{3}[/tex]

with [TEX]ab + bc + ca = abc = > \sum \frac {1}{a} = 1[/TEX]
[TEX]LHS = \sum \frac {\sqrt {b^2 + 2a^2}}{ab} = \sum \frac {\sqrt {b^2 + 2a^2}}{\sqrt {a^2b^2}} = \sum \sqrt {\frac {1}{a^2} + \frac {2}{b^2}}[/TEX]
by Mincopki
[TEX]LHS \ge \sqrt {(\sqrt {3})^2.(\sum \frac {1}{a})^2} = \sqrt {3}[/TEX]

 

[dandoh221]

Cho [tex]a,b,c>0;ab+bc+ca=abc[/tex]
CMR [tex]\sum \frac{sqrt{b^2+2a^2}}{ab} \geq \sqrt{3}[/tex]

đơn giản hơn 1 tí
[tex]\sum \frac{sqrt{b^2+2a^2}}{ab} \geq \sqrt{3}[/tex]
[TEX]\Leftrightarrow \sum \frac{sqrt{(b^2+2a^2)(1+2)}}{ab} \ge 3[/TEX]
[TEX]\sum \frac{sqrt{(b^2+2a^2)(1+2)}}{ab} \ge \sum \frac{b+2a}{ab} = 3\sum \frac{1}{a} = 3[/TEX]