Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(VT=\dfrac{\left(a+c\right)^2}{\left(a+c\right)\left(a+b\right)}+\dfrac{\left(b+d\right)^2}{\left(b+c\right)\left(b+d\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(c+d\right)}+\dfrac{\left(d+b\right)^2}{\left(d+a\right)\left(d+b\right)}\)
\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)
Dấu "=" xảy ra khi \(a=b=c=d\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+d}+\dfrac{d^2}{a+d}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\dfrac{\left(a+b+c+d\right)^2}{2\left(a+b+c+d\right)}=\dfrac{a+b+c+d}{2}=\dfrac{1}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)
Lời giải:
Áp dụng BĐT Cauchy- Schwarz:
\(\text{VT}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\geq \frac{(a+b+c+d)^2}{ab+bc+cd+da+2ac+2bd}\)
Lại có:
\((a+b+c+d)^2=[(a+c)+(b+d)]^2=(a+c)^2+(b+d)^2+2(a+c)(b+d)\)
Áp dụng BĐT Am-Gm:
\((a+c)^2+(b+d)^2\geq 4ac+4bd\)
\(\Rightarrow (a+b+c+d)^2\geq 4ac+4bd+2(ab+bc+cd+da)\)
\(\Rightarrow \text{VT}\geq \frac{(a+b+c+d)^2}{ab+bc+cd+da+2ac+2bd}\geq \frac{2(ab+bc+cd+da+2ac+2bd)}{ab+bc+cd+da+2ac+2bd}=2\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=d>0\)
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ba+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)(đúng)
a)Áp dụng
\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\left(1\right)\)
Lại có:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}=1\left(2\right)\)
Từ (1) và (2)=> đpcm
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có
\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
2)
Xét hiệu:
\(A^2+B^2+C^2+D^2+4-2A-2B-2C-2D\)
\(=\left(A^2-2A+1\right)+\left(B^2-2B+1\right)+\left(C^2-2C+1\right)+\left(D^2-2D+1\right)\)
\(=\left(A-1\right)^2+\left(B-1\right)^2+\left(C-1\right)^2+\left(D-1\right)^2\ge0\)
=> BĐT luôn đúng
Vậy \(A^2+B^2+C^2+D^2+4\ge2\left(A+B+C+D\right)\)
1)
Áp dụng BĐT Cauchy cho 2 số không âm, ta có:
\(\dfrac{AB}{C}+\dfrac{BC}{A}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{BC}{A}}=2B\) (1)
\(\dfrac{BC}{A}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{BC}{A}.\dfrac{AC}{B}}=2C\) (2)
\(\dfrac{AB}{C}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{AC}{B}}=2A\) (3)
Từ (1)(2)(3) cộng vế theo vế:
\(2\left(\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\right)\ge2\left(A+B+C\right)\)
\(\Rightarrow\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\ge A+B+C\)
\(\dfrac{a+b}{a+b+c}\)>\(\dfrac{a+b}{a+b+c+d}\)
\(\dfrac{b+c}{b+c+d}\)>\(\dfrac{b+c}{b+c+d+a}\)
\(\dfrac{c+d}{c+d+a}\)>\(\dfrac{c+d}{c+d+a+b}\)
\(\dfrac{d+a}{d+a+b}\)>\(\dfrac{d+a}{d+a+b+c}\)
cộng từng vế của bất đẳng thức lại với nhau ta được
\(\dfrac{a+b}{a+b+c}\)+\(\dfrac{b+c}{b+c+d}\)+\(\dfrac{c+d}{c+d+a}\)+\(\dfrac{d+a}{d+a+b}\)>\(\dfrac{a+b}{a+b+c+d}\)+\(\dfrac{b+c}{b+c+d+a}\)+\(\dfrac{c+d}{c+d+a+b}\)+\(\dfrac{d+a}{d+a+b+c}\)=\(\dfrac{2.\left(a+b+c+d\right)}{a+b+c+d}\)=2
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)
\(\geq \frac{(a+b+c+d)^2}{ab+ac+bc+bd+cd+ca+da+db}=\frac{(a+b+c+d)^2}{ab+cd+2ac+2bd+bc+da}\) (1)
Ta có:
\((a+b+c+d)^2=a^2+b^2+c^2+d^2+2ac+2bd+2(a+c)(b+d)\)
\(=a^2+b^2+c^2+d^2+2ac+2bd+2ab+2ad+2bc+2cd\)
Áp dụng BĐT AM-GM:
\(a^2+c^2\geq 2ac; b^2+d^2\geq 2bd\)
\(\Rightarrow (a+b+c+d)^2\geq 4ac+4bd+2ab+2ad+2bc+2cd\)
\(\Leftrightarrow (a+b+c+d)^2\geq 2(ab+cd+2ac+2bd+bc+da)\) (2)
Từ (1); (2) suy ra :
\(\text{VT}\geq \frac{2(ab+cd+2ac+2bd+bc+da)}{ab+cd+2ac+2bd+bc+da}=2\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=d\)