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Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
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
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}-\dfrac{a-b}{2\left(b+c\right)}+\dfrac{a-c}{2\left(b+c\right)}-\dfrac{a-c}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{2}\cdot\dfrac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{2}\left(\dfrac{1}{\left(b+c\right)\left(c+a\right)}-\dfrac{1}{\left(a+b\right)\left(b+c\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(a+c\right)\left(b+c\right)}\ge0\)(luôn đúng)
\(\Rightarrowđpcm\)
Áp dụng BĐT Bunhiacốpxki dạng phân thức có
\(\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge\dfrac{\left(a+b+c\right)^2}{a+2b^2+b+2c^2+c+2a^2}=\dfrac{9}{3+2\left(a^2+b^2+c^2\right)}\) (1)
Áp dụng BĐT Bunhiacốpxki có:
\(\left(a.1+b.1+c.1\right)^2\ge\left(1+1+1\right)\left(a^2+b^2+c^2\right)\)
\(\Rightarrow9\ge3\left(a^2+b^2+c^2\right)\Rightarrow3\ge a^2+b^2+c^2\Rightarrow2\left(a^2+b^2+c^2\right)\le6\) (2)
Thay (2) vào (1) có \(\dfrac{a^2}{a+b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+a^2}\ge\dfrac{9}{3+6}=1\) (đpcm)
Dấu = xảy ra khi a= b=c=1
5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
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 \(a=b\)
nhớ tìm trước khi hỏi Câu hỏi của Trần Huỳnh Cẩm Hân - Toán lớp 8 | Học trực tuyến