2a/b+c+b+c/2a≥2(a,b,c>0)
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b. \(\left(a+c\right)\left(a-c\right)-b\left(2a-b\right)-\left(a-b+c\right)\left(a-b-c\right)\)
=\(\left(a^2-c^2\right)-2ab+b^2-\left(a-b\right)^2+c^2\)
=\(a^2-c^2-2ab+b^2-a^2+2ab-b^2+c^2\)
=0=VP=> đpcm
a. \(\left(a-1\right)\left(a-2\right)+\left(a-3\right)\left(a+4\right)-\left(2a^2+5a-34\right)\)
=\(\left(a^2-3a+2\right)+\left(a^2+a-12\right)-\left(2a^2+5a-34\right)\)
=\(a^2-3a+2+a^2+a-12-2a^2-5a+34\)
=-7a+24
=VP => đpcm
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\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+1a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)
\(\Leftrightarrow2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2c}\right)\ge1+\dfrac{b+2a}{b+2a}+\dfrac{c+2b}{c+2b}+\dfrac{a+2c}{a+2c}=1+1+1+1=4\)Thật vậy:
\(\dfrac{a}{b+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2b}+\dfrac{c}{a+2c}=a\left(\dfrac{1}{b+2c}+\dfrac{1}{b+2a}\right)+b\left(\dfrac{1}{c+2a}+\dfrac{1}{c+2b}\right)+c\left(\dfrac{1}{a+2b}+\dfrac{1}{a+2c}\right)\)
\(\ge\dfrac{4a}{2\left(a+b+c\right)}+\dfrac{4b}{2\left(a+b+c\right)}+\dfrac{4c}{2\left(a+b+c\right)}=2\)
\(\Rightarrow VT\ge2.2=4\)
\(\RightarrowĐPCM\)
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Từ 2a + b + c = 0 <=> a + a + b + c = 0 <=> a + c = -(a + b)
Ta có: VT = 2a3 + b3 + c3 = (a3 + b3) + (a3 + c3)
= (a + b)(a2 - ab + b2) + (a + c)(a2 - ac + c2)
= (a + b)(a2 + 2ab + b2) - 3ab(a + b) + (a + c)(a2 + 2ac + c2) - 3ac(a + c)
= (a + b)3 - 3ab(a + b) + (a + c)3 - 3ac(a + c)
= (a + b)3 - (a + b)3 - 3ab(a + b) + 3ac(a + b)
= -3a(a + b)(b - c) = 3a(a + b)(c - b) = VP
=> VT = VP => đpcm
Áp dụng bđt cauchy, ta có
\(\dfrac{2a}{b+c}+\dfrac{b+c}{2a}\ge2.\sqrt{\dfrac{2a}{b+c}\dfrac{b+c}{2a}}=2\)(đpcm)
Dấu ''='' xảy ra <=>2a = b + c