Cho a,b,c là độ dài 3 cạnh của tam giác có p = \(\dfrac{a+b+c}{2}\)
CMR : \(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}>2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
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\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Giả sử đpcm là đúng , khi đó , ta có :
\(\left|\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\right)< 1\right|\)
\(\Leftrightarrow\left|\frac{a-c}{b}+\frac{b-a}{c}+\frac{c-b}{a}\right|< 1\)
\(\Leftrightarrow\left|\frac{\left(a-c\right)ac+\left(b-a\right)ab+\left(c-b\right)bc}{abc}\right|< 1\)
Lại có : \(\left(a-c\right)ac+\left(b-a\right)ab+\left(c-b\right)bc\)
\(=\left(a-c\right)ac-\left(a-c+c-b\right)ab+\left(c-b\right)bc\)
\(=\left(a-c\right)\left(ac-ab\right)-\left(c-b\right)\left(ab-bc\right)\)
\(=a\left(a-c\right)\left(c-b\right)-b\left(c-b\right)\left(a-c\right)\)
\(=\left(a-c\right)\left(c-b\right)\left(a-b\right)\)
\(\Rightarrow\left|\frac{\left(a-c\right)\left(c-b\right)\left(a-b\right)}{abc}\right|< 1\) ( 1 )
Mặt khác : a ; b ; c là 3 cạnh tam giác
=> \(\frac{\left|a-c\right|}{b}< 1;\frac{\left|b-a\right|}{c}< 1;\frac{\left|c-b\right|}{a}< 1\)
\(\Rightarrow\frac{\left|\left(a-c\right)\left(b-a\right)\left(c-b\right)\right|}{abc}< 1\) ( 2 )
Biểu thức trong giá trị tuyệt đối của ( 1 ) ; ( 2 ) đối nhau
=> từ ( 2 ) => (1)
=> Điều giả sử là đúng
=> ĐPCM
a)a,b,c là độ dài 3 cạnh của 1 tam giác
\(\Rightarrow a< b+c\Rightarrow a^2< ab+ac\)
TT\(\Rightarrow b^2< ba+bc\)
\(c^2< ca+cb\)
Cộng vế theo vế ta có đpcm
b)BĐT\(\Leftrightarrow\dfrac{a}{b+c-a}+\dfrac{1}{2}+\dfrac{b}{a+c-b}+\dfrac{1}{2}+\dfrac{c}{a+b-c}+\dfrac{1}{2}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{a+b+c}{b+c-a}+\dfrac{a+b+c}{a+c-b}+\dfrac{a+b+c}{a+b-c}\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\right)\ge9\)(đúng theo AM-GM)
Ta có :
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{2}{c}\)
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{p-a+p-c}=\dfrac{2}{a}\)
\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{p-c+p-a}=\dfrac{2}{b}\)
Cộng từng về ta có đpcm
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\right)\)
Áp dụng:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{4}{2p-a-b}\)
Mà \(2p=a+b+c\)
\(\Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{a+b+c-a-b}=\dfrac{4}{c}\)
Tương tự \(\Rightarrow2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\)
\(\Rightarrowđpcm\)
Vì vai trò của a,b,c là như nhau, giả sử
\(a\ge c\ge b>0\)
Ta có
\(a+b-c< a\)
\(\Leftrightarrow b-c\le0\) ( đúng với gt )
\(\Rightarrow a+b-c< a\)
\(\Leftrightarrow\left(a+b-c\right)^2< a^2\)
\(\Leftrightarrow\dfrac{1}{\left(a+b-c\right)^2}\ge\dfrac{1}{a^2}\)
CMTT :
\(\dfrac{1}{\left(b+c-a\right)^2}\ge\dfrac{1}{b^2};\dfrac{1}{\left(c+a-b\right)^2}\ge\dfrac{1}{c^2}\)
Cộng vế với vế 3 BĐT trên , được
\(\dfrac{1}{\left(a+b-c\right)^2}+\dfrac{1}{\left(b+c-a\right)^2}+\dfrac{1}{\left(c+a-b\right)^2}\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
B1:
\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
Xét hiệu:
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
=> BĐT luôn đúng
*
Ta có:
\(a< b+c\Rightarrow a^2< ab+ac\)
\(b< a+c\Rightarrow b^2< ab+ac\)
\(c< a+b\Rightarrow a^2< ac+bc\)
Cộng từng vế bất đẳng thức ta được:
\(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
Vậy: \(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
B2:
Ta có: \(a+b>c\) ; \(b+c>a\); \(a+c>b\)
Xét:\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{a+b+c}+\dfrac{1}{a+c+b}=\dfrac{2}{a+b+c}>\dfrac{2}{b+c+b+c}=\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+c+a+c}=\dfrac{1}{a+c}\)
Suy ra:
\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
=> ĐPCM
Đề phải là \(\ge\)
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}=\dfrac{1}{\dfrac{-a+b+c}{2}}+\dfrac{1}{\dfrac{a-b+c}{2}}+\dfrac{1}{\dfrac{a+b-c}{2}}=2\left(\dfrac{1}{-a+b+c}+\dfrac{1}{a-b+c}+\dfrac{1}{a+b-c}\right)\)
Áp dụng BĐT trong tam giác:
a+b>c=>a+b-c>0
a+c>b=>a-b+c>0
b+c>a=>-a+b+c>0
Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)cho 2 số dương:
\(\dfrac{1}{-a+b+c}+\dfrac{1}{a-b+c}\ge\dfrac{4}{2c}=\dfrac{2}{c}\)
Dấu = xảy ra khi -a+b+c=a-b+c<=>a=b
\(\dfrac{1}{a-b+c}+\dfrac{1}{a+b-c}\ge\dfrac{4}{2a}=\dfrac{2}{a}\)
Dấu = xảy ra khi a-b+c=a+b-c<=>b=c
\(\dfrac{1}{a+b-c}+\dfrac{1}{-a+b+c}\ge\dfrac{4}{2b}=\dfrac{2}{b}\)
Dấu = xảy ra khi a+b-c=-a+b+c<=>a=c
=>\(2\left(\dfrac{1}{-a+b+c}+\dfrac{1}{a-b+c}+\dfrac{1}{a+b-c}\right)\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
Hay \(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)<=>tam giác ABC đều