cho a,b,c là 3 cạnh của tam giác. Chứng minh rằng:
a) \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< 2\)
b)\(a^3+b^3+c^3+3abc>ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)
Cho a,b,c dương. Chứng minh
\(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(c+a\right)^2}\ge\dfrac{3\sqrt{3abc\left(a+b+c\right)}.\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)^3}\)
Xét:
\(\dfrac{c}{a-b}.\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}=1+\dfrac{c}{a-b}.\dfrac{c\left(a-b\right)-\left(a^2-b^2\right)}{ab}=1+\dfrac{c}{a-b}.\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab}=1+\dfrac{c^2-c\left(a+b\right)}{ab}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)
CMTT cộng theo vế:
\(BTCCM=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}=\dfrac{6\left(a^3+b^3+c^3\right)}{3abc}\)
Mà Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) ( tự cm,ez)
Vậy \(BTCCM=3+6=9\left(đpcm\right)\)
Cho tam giác ABC có BC=a,AC=b,AB=c. Chứng minh rằng: \(3\left(a^3+b^3+c^3\right)+4abc\ge\dfrac{13}{27}\left(a+b+c\right)^3\)
Do a;b;c là độ dài 3 cạnh của tam giác
\(\Rightarrow abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)
\(\Leftrightarrow4\left(a^3+b^3+c^3\right)+15abc\ge\left(a+b+c\right)^3\)
\(\Leftrightarrow3\left(a^3+b^3+c^3\right)+\dfrac{45}{4}abc\ge\dfrac{3}{4}\left(a+b+c\right)^3\)
\(\Rightarrow3\left(a^3+b^3+c^3\right)+4abc\ge\dfrac{3}{4}\left(a+b+c\right)^3-\dfrac{29}{4}abc\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{3}{4}\left(a+b+c\right)^3-\dfrac{29}{4}abc\ge\dfrac{13}{27}\left(a+b+c\right)^3\)
\(\Leftrightarrow\left(a+b+c\right)^3\ge27abc\) (hiển nhiên đúng theo AM-GM)
Cho 3 số a, b, c thỏa mãn a # -b, b # -c, c # -a.
Chứng minh rằng : \(\dfrac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^2-ab}{\left(c+a\right)\left(c+b\right)}=0\)
chứng minh rằng
a) \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
b)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\cdot\left(a^2+b^2+c^2+ab+bc-ca\right)\)
áp dụng suy ra kết quả
a) \(a^3+b^3+c^3=3abc\) thì \(\left\{{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
b) cho \(a^3+b^3+c^3=3abc\left(a+c\ne0\right)\)
tính B= \(\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
Câu 1:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
cho a,b,c là số thực dương chứng minh
\(\dfrac{2\left(a^4+b^4+c^4\right)}{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}+\dfrac{ab+bc+ca}{a^3+b^3+c^3}\ge2\)
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Cho \(a;b;c\) là các số thực dương thỏa mãn :\(0< a;b;c< 1\). Chứng minh rằng:
\(\dfrac{1}{a.\left(1-b\right)}+\dfrac{1}{b.\left(1-c\right)}+\dfrac{1}{c.\left(1-a\right)}\ge\dfrac{3}{1-\left(a+b+c\right)+ab+bc+ac}\)
P/s: Đề cương toán lớp 10 trường THPT chuyên sư phạm Hà Nội.
Em xin nhờ quý thầy cô giáo và các bạn giúp đỡ, em cám ơn nhiều ạ!
Đặt \(a\left(1-b\right)=x;b\left(1-c\right)=y;c\left(1-a\right)=x\)
\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca=1-a\left(1-b\right)-b\left(1-c\right)-c\left(1-a\right)=1-x-y-z\)
BĐT cần c/m trở thành:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{3}{1-x-y-z}\)
\(\Leftrightarrow\left(1-x-y-z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-3\ge0\)
\(\Leftrightarrow\dfrac{1-x-y-z}{x}+\dfrac{1-x-y-z}{y}+\dfrac{1-x-y-z}{z}-3\ge0\)
\(\Leftrightarrow\dfrac{1-y-z}{x}+\dfrac{1-z-x}{y}+\dfrac{1-x-y}{z}-6\ge0\) (1)
Lại có: \(1-y-z=1-b\left(1-c\right)-c\left(1-a\right)=1-b-c+bc+ca=\left(1-b\right)\left(1-c\right)+ca\)
Nên (1) tương đương:
\(\dfrac{\left(1-b\right)\left(1-c\right)+ca}{a\left(1-b\right)}+\dfrac{\left(1-a\right)\left(1-c\right)+ab}{b\left(1-c\right)}+\dfrac{\left(1-a\right)\left(1-b\right)+bc}{c\left(1-a\right)}-6\ge0\)
\(\Leftrightarrow\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\)
BĐT trên hiển nhiên đúng theo AM-GM do:
\(\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\sqrt[6]{\dfrac{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}}=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
cho a,b,c là các số thực dương. Chứng minh rằng :
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(a+b+c\right)\)
AD bđt AM-GM cho 3 số
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)
\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)
thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)