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Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$
$\Leftrightarrow a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}=\frac{(\frac{3}{2})^2}{3}=\frac{3}{4}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=\frac{1}{2}$.
\(\Leftrightarrow\dfrac{2a^2}{b^2}+\dfrac{2b^2}{c^2}+\dfrac{2c^2}{a^2}=\dfrac{2a}{c}+\dfrac{2c}{b}+\dfrac{2b}{a}\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
Ta có:(A1)\(^2\)\(\ge\)0
\(\Leftrightarrow a^2-a+\dfrac{1}{4}\ge0\\ \Leftrightarrow a^2+\dfrac{1}{4}\ge a\left(1\right)\\ cmtt:b^2+\dfrac{1}{4}\ge b\left(2\right)\\ 6^2+\dfrac{1}{4}\ge c\left(3\right)\)
Cộng (1);(2) và (3) theo vế, ta có:
\(a^2+\dfrac{1}{4}+b^2+\dfrac{1}{4}+6^2+\dfrac{1}{4}\ge a+b+c\\ \Leftrightarrow a^2+b^2+c^2+\dfrac{3}{4}\ge\dfrac{3}{2}\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{2}-\dfrac{3}{4}\\ \Leftrightarrow a^2+b^2+c^2\ge\dfrac{3}{4}\)
\(\left(a+b+c\right)^2=\dfrac{9}{4}\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=\dfrac{9}{4}\)
Có \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(b^2+c^2\ge2\sqrt{b^2c^2}=2bc\)
\(a^2+c^2\ge2\sqrt{a^2c^2}=2ac\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc\le a^2+b^2+c^2+a^2+b^2+a^2+c^2+b^2+c^2=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{9}{4}\le3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^2+b^2+c^2\ge\dfrac{9}{4}.\dfrac{1}{3}=\dfrac{3}{4}\left(ĐPCM\right)\)
Bài này áp dụng BĐT cosi nha bn
A/dụng bđt bunhiacopxki có:
\(\left(a+b+c\right)^2\le\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\left(\dfrac{3}{2}\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{9}{4}:3=\dfrac{3}{4}\)(đpcm)
Dấu ''='' xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Bài 1:a,b,c ba cạnh tam giác => a,b,c dương
\(\left\{{}\begin{matrix}a+c>b\\a+b>c\\b+c>a\end{matrix}\right.\) ta có: \(\dfrac{x}{y}< \dfrac{x+p}{y+p}\forall_{x,y,p>0\&x< y}\)
\(VT=\dfrac{a}{a+b}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+c}{a+b}+\dfrac{b}{c+a}< \dfrac{a+c+c}{a+b+c}+\dfrac{b+b}{a+b+c}=\)
\(=\dfrac{a+b+c+b+c}{a+b+c}< \dfrac{\left(a+b+c\right)+\left(A+b+c\right)}{a+b+c}< \dfrac{2\left(b+a+c\right)}{a+b+c}=2=VP\)
p/s: đề sao làm vậy:
mình nghi đề phải thế này: \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) cách làm đơn giản hơn
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 \(\Leftrightarrow a=b=c=1\)
\(a^2+\dfrac{1}{4}>=2\cdot a\cdot\dfrac{1}{2}=a\)
\(b^2+\dfrac{1}{4}>=b\)
c^2+1/4>=c
=>a^2+1/4+b^2+1/4+c^2+1/4>=a+b+c
=>a^2+b^2+c^2>=3/4