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e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
Bài 1:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0
Áp dụng BĐT Chauchy cho 2 số không âm, ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)
Cộng vế theo vế ta được:
\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
3) Biến đổi tương đương:
\(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3\) (1)
\(\Leftrightarrow\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+6\left(a^3+c^3+b^3\right)\)
\(\ge\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)
\(\Leftrightarrow\left[a^3+b^3-ab\left(a+b\right)\right]+\left[a^3+c^3-ac\left(a+c\right)\right]+\left[b^3+c^3-bc\left(b+c\right)\right]\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(a+c\right)\left(a-c\right)^2+\left(b+c\right)\left(b-c\right)^2\ge0\) luôn đúng do a, b, c > 0
=> (1) đúng
Dấu "=" xảy ra khi a = b = c
4) Ta có: a+b>c ; b+c>a; a+c>b
Xét \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
Tương tự: \(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy suy ra được điều phải chứng minh
a.
\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)
\(\Leftrightarrow a^4+b^4\ge ab^3+a^3b\)
\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(*)
Mà \(a^2+ab+b^2=\left(a^2+2\cdot a\cdot\dfrac{1}{2}b+\dfrac{b^2}{4}\right)+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\)
Suy ra (*) đúng => đpcm
Dấu "=" xảy ra khi a = b
b.
\(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(\Leftrightarrow3a^4+3b^4+3c^4\ge a^4+ab^3+ac^3+a^3b+b^4+bc^3+a^3c+b^3c+c^4\)
\(\Leftrightarrow2a^4+2b^4+2c^4\ge ab^3+a^3b+b^3c+bc^3+ca^3+c^3a\)
\(\Leftrightarrow\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge\left(a^3b+ab^3\right)+\left(b^3c+bc^3\right)+\left(c^3a+ca^3\right)\)
Theo câu a. thì điều này đúng
Dấu "=" khi a=b=c
Lời giải:
Sử dụng pp biến đổi tương đương:
a) \(\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2\)
\(\Leftrightarrow \frac{a^2+b^2}{2}\geq \frac{(a+b)^2}{4}\)
\(\Leftrightarrow 4(a^2+b^2)\geq 2(a+b)^2\Leftrightarrow 4(a^2+b^2)\geq 2(a^2+2ab+b^2)\)
\(\Leftrightarrow 2(a^2+b^2)\geq 4ab\Leftrightarrow 2(a^2+b^2-2ab)\geq 0\)
\(\Leftrightarrow 2(a-b)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xẩy ra khi $a=b$
c)
\(\frac{a^2+b^2+c^2}{3}\geq \left(\frac{a+b+c}{3}\right)^2\) \(\Leftrightarrow \frac{a^2+b^2+c^2}{3}\geq \frac{(a+b+c)^2}{9}\)
\(\Leftrightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\)
\(\Leftrightarrow 3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)\)
\(\Leftrightarrow 2(a^2+b^2+c^2)\geq 2(ab+bc+ac)\)
\(\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)\geq 0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b=c$
b) \(\frac{a^4+b^4}{2}\geq \left(\frac{a+b}{2}\right)^4\)
Áp dụng 2 lần BĐT phần a: \(\frac{a^4+b^4}{2}\geq \left(\frac{a^2+b^2}{2}\right)^2(1)\)
Và: \(\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2\Rightarrow \left(\frac{a^2+b^2}{2}\right)^2\geq \left(\frac{a+b}{2}\right)^4(2)\)
Từ \((1); (2)\Rightarrow \frac{a^4+b^4}{2}\geq \left(\frac{a+b}{2}\right)^4\) (đpcm)
Dấu bằng xảy ra khi \(a=b\)
Áp dụng BĐT AM - GM, ta có:
\(\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\)
\(\ge\dfrac{3\sqrt{a^3b^3c^3}}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{a^2+c^2}{b^2+\dfrac{a^2+c^2}{2}}\)
\(\ge\dfrac{3abc}{2abc}+\dfrac{2\left(a^2+b^2\right)}{2c^2+a^2+b^2}+\dfrac{2\left(b^2+c^2\right)}{2a^2+b^2+c^2}+\dfrac{2\left(a^2+c^2\right)}{2b^2+a^2+c^2}\)
\(=\dfrac{3}{2}+2\times\left[\dfrac{a^2+b^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\dfrac{b^2+c^2}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\dfrac{c^2+a^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\right]\) (1)
Đặt \(\left\{{}\begin{matrix}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{matrix}\right.\), ta có:
\(\left(1\right)\Leftrightarrow\dfrac{3}{2}+2\times\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\)
\(\ge\dfrac{3}{2}+2\times\dfrac{3}{2}\) (Bất_đẳng_thức_Nesbitt)
\(=\dfrac{9}{2}\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c
AM-GM ngược dấu như sau:
\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3ab}=\dfrac{2a-b}{3}\)
Tương tự ta cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{2b-c}{3};\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{2c-a}{3}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a-b}{3}+\dfrac{2b-c}{3}+\dfrac{2c-a}{3}=\dfrac{a+b+c}{3}=VP\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)
Dễ thấy :
\(a^{3}+b^{3}+c^{3}+ab(b+c)+bc(b+c)+ca(c+a)=(a^{2}+ b^{2}+c^{2})(a+b+c)\)
\(\Rightarrow VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
Vậy cần chứng minh
\(\dfrac{a^2+b^2+c^2}{a+b+c}\ge\dfrac{a+b+c}{3}\Leftrightarrow\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2\right)\) (luôn đúng)