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Lời giải:
Áp dụng BĐT AM-GM cho các số dương ta có:
\(\frac{a^3}{(a+1)(b+1)}+\frac{a+1}{8}+\frac{b+1}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(b+1)(c+1)}+\frac{b+1}{8}+\frac{c+1}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(c+1)(a+1)}+\frac{c+1}{8}+\frac{a+1}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng theo vế:
\(\Rightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{a+b+c+3}{4}\geq \frac{3}{4}(a+b+c)\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{3}{2}\geq \frac{9}{4}\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) và BĐT cần chứng minh là:
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel và AM-GM ta có:
\(VT=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}=VP\)
Xảy ra khi \(x=y=z=1 \Rightarrow a=b=c=1\)
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nhớ giải câu hỏi nhé ! thanks
Ta có \(\dfrac{1}{a^3\left(b+c\right)}=\dfrac{1}{\dfrac{1}{b^3c^3}\left(b+c\right)}=\dfrac{b^2c^2}{\dfrac{1}{b}+\dfrac{1}{c}}\)
Tương tự \(\Rightarrow VT=\dfrac{b^2c^2}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{c^2a^2}{\dfrac{1}{c}+\dfrac{1}{a}}+\dfrac{a^2b^2}{\dfrac{1}{a}+\dfrac{1}{b}}\)
\(\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}\) (BĐT B.C.S)
\(=\dfrac{\left(ab+bc+ca\right)^2}{2\left(\dfrac{ab+bc+ca}{abc}\right)}\)
\(=\dfrac{ab+bc+ca}{2}\) (do \(abc=1\))
\(\ge\dfrac{3\sqrt[3]{abbcca}}{2}\)
\(=\dfrac{3\left(\sqrt[3]{abc}\right)^2}{2}=\dfrac{3}{2}\) (do \(abc=1\))
ĐTXR \(\Leftrightarrow a=b=c=1\)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Ta có: \(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}=-1\)
Trường hợp 2: a=b=c
\(\Leftrightarrow N=\dfrac{b+b}{b}\cdot\dfrac{a+a}{a}\cdot\dfrac{c+c}{c}=2\cdot2\cdot2=8\)
1, Ta có a^3+b^3+c^3=3abc
-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2
-> (a+b)3 + c^3 - 3ab(a+b+c)=0
-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0
-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0
Th1: a+b+c=0
->P= a+b/2 . b+c/2 . c+a/2
= (-c)(-a)(-b)/2=-1
TH2 a^2+b^2+c^2-ab-bc-ca=0
->2a^2+2b^2+2c^2-2ab-abc-2ac=0
->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
-> (a-b)^2+(a-c)^2+(b-c)^2=0
Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0
Dấu = xảy ra (=)a-b=0
b-c=0
a-c=0
-> a=b=c
->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8
-Hoặc có thể đề bạn sai
hiểu đơn giản thì có thể hiểu như sau:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc>abc\)
-Không có a;b;c thỏa mãn
Lời giải:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b).\frac{ab+c(a+b+c)}{abc(a+b+c)}=0\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\)
\(\Rightarrow (a+b)(c+a)(c+b)=0\)
Do đó:
\(A=(a^3+b^3)(b^3+c^3)(c^3+a^3)\)
\(=(a+b)(a^2-ab+b^2)(b+c)(b^2-bc+c^2)(c+a)(c^2-ca+a^2)\)
\(=(a+b)(c+a)(c+b)[(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)]=0\)