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\(\Leftrightarrow ab\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+bc\left(\dfrac{1}{a+c}-\dfrac{1}{a+b}\right)+ca\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)=0\)
\(\Leftrightarrow\dfrac{ab\left(a-b\right)}{\left(b+c\right)\left(a+c\right)}+\dfrac{bc\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{ca\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\dfrac{ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\) hay tam giác cân
đặt\(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{x+z}{2}\\b=\dfrac{x+y}{2}\\c=\dfrac{y+z}{2}\end{matrix}\right.\)
sau đó thay vào bt rồi tính là ra
159. Cho ba số dương a,b,c. Chứng minh: \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
Áp dụng bất đẳng thức \(a^2+b^2+c^2\ge ab+bc+ca\) có:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{a^2b}{b}+\dfrac{b^2c}{c}+\dfrac{c^2a}{a}\)
\(=a^2+b^2+c^2\ge ab+bc+ca\)
Dấu " = " khi a = b = c = 1
Vậy...
Lời giải:
\(\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c\)
\(\Leftrightarrow \frac{a^2+bc}{b+c}-c+\frac{b^2+ac}{a+c}-a+\frac{c^2+ab}{a+b}-b\geq 0\)
\(\Leftrightarrow \frac{a^2-c^2}{b+c}+\frac{b^2-a^2}{a+c}+\frac{c^2-b^2}{a+b}\geq 0\)
\(\Leftrightarrow a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\geq 0\)
\(\Leftrightarrow \frac{a^2(a-b)(a+b)+b^2(b-c)(b+c)+c^2(c-a)(c+a)}{(a+b)(b+c)(c+a)}\geq 0\)
\(\Leftrightarrow a^2(a^2-b^2)+b^2(b^2-c^2)+c^2(c^2-a^2)\geq 0\)
\(\Leftrightarrow a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)\geq 0\)
\(\Leftrightarrow \frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{2}\geq 0\) (luôn đúng)
Do đó ta có đpcm
Dấu bằng xảy ra khi $a=b=c$
1) 2( a2 + b2 ) ≥ ( a + b)2
<=> 2a2 + 2b2 - a2 - 2ab - b2 ≥ 0
<=> a2 - 2ab + b2 ≥ 0
<=> ( a - b )2 ≥ 0 ( luôn đúng )
=> đpcm
2) Áp dụng BĐT Cô-si cho 2 số dương x , y , ta có :
a + b ≥ \(2\sqrt{ab}\)
=> \(\dfrac{1}{x}+\dfrac{1}{y}\) ≥ 2\(\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\)
=> ( x + y)( \(\dfrac{1}{x}+\dfrac{1}{y}\) ) ≥ \(2\sqrt{xy}\)2\(\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\)
=> ( x + y)( \(\dfrac{1}{x}+\dfrac{1}{y}\)) ≥ 4
=> \(\dfrac{1}{x}+\dfrac{1}{y}\) ≥ \(\dfrac{4}{x+y}\)
\(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}=\dfrac{2}{b}\)
Tương tự:
\(\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{2}{a}\) ; \(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c}\)
Cộng vế:
\(2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
cho em hỏi tại sao 1/a+b-c +1/b+c-a>=4/a+b-c+b+c-a vậy ạ
+ 
+ 
≥ 3.
Đặt b + c – a = x > 0 (1); a + c – b = y > 0 (2); a + b – c = z > 0 (3)
Cộng (1) và (2) => b + c – a + a + c – b = x + y ⇔ 2c = x + y ⇔ c = 
Tương tự a = 
; b = 
Do đó + + = 
+ 
+ 
= 
(
+ 
+ 
+ 
+ 
+ 
)
= [( + ) + ( + ) + ( + )] ≥ (2 + 2 + 2) = 3.
Vậy + + ≥ 3.
Gọi cái đó là P
Đặt \(\left\{{}\begin{matrix}b+c-a=2x\\c+a-b=2y\\a+b-c=2z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=y+z\\b=z+x\\c=x+y\end{matrix}\right.\)
Thì ta có:
\(P=\dfrac{\left(x+z\right)\left(y+z\right)}{2z}+\dfrac{\left(x+y\right)\left(z+y\right)}{2y}+\dfrac{\left(z+x\right)\left(y+x\right)}{2x}\ge2\left(x+y+z\right)\)
\(\Leftrightarrow2x^2y^2+2y^2z^2+2z^2x^2-2xyz^2-2yzx^2-2zxy^2\ge0\)
\(\Leftrightarrow\left(xy-yz\right)^2+\left(yz-zx\right)^2+\left(zx-xy\right)^2\ge0\) (đúng)
\(\RightarrowĐPCM\)