Question

6. Bất đẳng thức

Bài 9: Cho a, b, c, d, e \(\in\) R. Chứng minh các bất đẳng thức sau:

a. \(a^2+b^2+c^2\ge ab+bc+ca\)

b. \(a^2+b^2+1\ge ab+a+b\)

c. \(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

d. \(a^2+b^2+c^2\ge2\left(ab+bc-ca\right)\)

e. \(a^4+b^4+c^2+1\ge2a\left(ab^2-a+c+1\right)\)

f. \(\frac{a^2}{4}+b^2+c^2\ge ab-ac+2bc\)

g. \(a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\ge6abc\)

h. \(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)

i. \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\) với a, b, c >0

k. \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\) với a, b, c \(\ge\)0

Answer 1

a.

\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

(luôn đúng)

b. Áp dụng BĐT \(x^2+y^2\ge2xy\)

\(a^2+b^2\ge2ab,a^2+1\ge2a,b^2+1\ge2b\)\(\Rightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\Leftrightarrow a^2+b^2+1\ge ab+a+b\)

c. Tương tự câu b

Answer 2

Áp dụng BĐT Cô si ta có

i. \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},\frac{1}{b}+\frac{1}{c}\ge\frac{2}{\sqrt{bc}},\frac{1}{c}+\frac{1}{a}\ge\frac{2}{\sqrt{ca}}\)

\(\Rightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)

k. Tương tự câu i