Cho a,b,c là độ dài 3 cạnh một tam giác. Tìm:
\(MinP=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}\)
Cho a,b,c là độ dài 3 cạnh của 1 tam giác. Tìm: \(MinP=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}\)
Cho \(a,b,c>0\) thỏa mãn \(ab+bc+ca=3\).Tìm Max:
\(P=\dfrac{a}{a^2+4a+3}+\dfrac{b}{b^2+4b+3}+\dfrac{c}{c^2+4c+3}\)
\(ab+bc+ca=3\Rightarrow\left\{{}\begin{matrix}a+b+c\ge3\\abc\le1\end{matrix}\right.\)
Ta sẽ chứng minh \(P\le\dfrac{3}{8}\)
\(P\le\dfrac{a}{6a+2}+\dfrac{b}{6b+2}+\dfrac{c}{6c+2}\) nên chỉ cần chứng minh: \(\dfrac{a}{3a+1}+\dfrac{b}{3b+1}+\dfrac{c}{3c+1}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{3a+1}+\dfrac{1}{3b+1}+\dfrac{1}{3c+1}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{\left(3a+1\right)\left(3b+1\right)+\left(3b+1\right)\left(3c+1\right)+\left(3c+1\right)\left(3a+1\right)}{\left(3a+1\right)\left(3b+1\right)\left(3c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{6\left(a+b+c\right)+30}{27abc+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}\)
\(\Rightarrow\dfrac{6\left(a+b+c\right)+30}{27+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}\)
\(\Leftrightarrow24\left(a+b+c\right)+120\ge165+9\left(a+b+c\right)\)
\(\Leftrightarrow a+b+c\ge3\) (đúng)
Cho tam giác ABC có chu vi bằng 2. Ký hiệu a, b, c là độ dài ba cạnh của tam giác.
Tìm giá trị nhỏ nhất của biểu thức \(\dfrac{a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{9c}{a+b-c}\)
Đặt b + c - a = x; c + a - b = y; a + b - c = z. (x, y, z > 0)
Ta có \(A=\dfrac{a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{9c}{a+b-c}=\dfrac{y+z}{2x}+\dfrac{2\left(z+x\right)}{y}+\dfrac{9\left(x+y\right)}{2z}=\left(\dfrac{y}{2x}+\dfrac{2x}{y}\right)+\left(\dfrac{z}{2x}+\dfrac{9x}{2z}\right)+\left(\dfrac{9y}{2z}+\dfrac{2z}{y}\right)\ge2\sqrt{\dfrac{y}{2x}.\dfrac{2x}{y}}+2\sqrt{\dfrac{z}{2x}.\dfrac{9x}{2z}}+2\sqrt{\dfrac{9y}{2z}.\dfrac{2z}{y}}=2+3+6=11\).
Dấu "=" xảy ra khi và chỉ khi \(3y=2z=6x\Leftrightarrow3\left(c+a-b\right)=2\left(b+c-a\right)=6\left(a+b-c\right)\)
\(\Leftrightarrow a=\dfrac{5}{6};b=\dfrac{2}{3};c=\dfrac{1}{2}\).
cho a, b, c là số đo độ dài 3 cạnh một tam giác
chứng minh \(\dfrac{b+c}{b+-a}+\dfrac{c+a}{c+a-b}+\dfrac{a+b}{a+b-c}\ge6\)
Ta có bất đẳng thức sau
a2 + b2 + c2 \(\ge\) ab + bc + ca (1)
Dấu "=" xảy ra <=> a = b = c
Thật vậy (1) <=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca \(\ge0\)
<=> (a - b)2 + (b - c)2 + (c - a)2 \(\ge0\) (bđt này luôn đúng)
Khi đó ta được (1) <=> 2(a2 + b2 + c2) \(\ge\) 2(ab + bc + ca)
<=> 3(a2 + b2 + c2) \(\ge\) 2ab + 2bc + 2ca + a2 + b2 + c2
<=> 3(a2 + b2 + c2) \(\ge\) (a + b + c)2
=> -(a2 + b2 + c2) \(\le\dfrac{(a+b+c)^2}{3}\)
Ta có \(P=\dfrac{b+c}{b+c-a}+\dfrac{c+a}{c+a-b}+\dfrac{a+b}{a+b-c}\)
\(=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}+3\)
\(=\dfrac{a^2}{ab+ac-a^2}+\dfrac{b^2}{ab+bc-b^2}+\dfrac{c^2}{ac+bc-c^2}+3\)
\(\ge\dfrac{\left(a+b+c\right)^2}{ab+ac-a^2+ab+bc-b^2+ac+bc-c^2}+3\) (BĐT Schwarz)
\(=\dfrac{\left(a+b+c\right)^2}{2ab+2ac+2bc-a^2-b^2-c^2}+3\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2-2\left(a^2+b^2+c^2\right)}+3\)
\(\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2-\dfrac{2}{3}\left(a+b+c\right)^2}+3=\dfrac{1}{1-\dfrac{2}{3}}+3=6\) (đpcm)
Cho a,b,c là độ dài 3 cạnh của một tam giác. Chứng minh \(\dfrac{1}{a+b-c}\)+\(\dfrac{1}{b+c-a}\)+\(\dfrac{1}{c+a-b}\)≥\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)
Mọi người giúp mình nhé
\(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}=\dfrac{2}{b}\)
Tương tự:
\(\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{2}{a}\) ; \(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c}\)
Cộng vế:
\(2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Cho tam giác ABC có độ dài 3 cạnh là a, b, c thỏa mãn: \(\dfrac{ab}{b+c}+\dfrac{bc}{c+a}+\dfrac{ac}{a+b}=\dfrac{ac}{b+c}+\dfrac{ab}{c+a}+\dfrac{bc}{a+b}\). Chứng minh: Tam giác ABC cân
Cho tam giác ABC có độ dài 3 cạnh là a, b, c thỏa mãn: \(\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{ac}{a+b}=\dfrac{ac}{b+c}+\dfrac{ab}{a+c}+\dfrac{bc}{a+b}\). Chứng minh tam giác ABC cân
\(\Leftrightarrow ab\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+bc\left(\dfrac{1}{a+c}-\dfrac{1}{a+b}\right)+ca\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)=0\)
\(\Leftrightarrow\dfrac{ab\left(a-b\right)}{\left(b+c\right)\left(a+c\right)}+\dfrac{bc\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{ca\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\dfrac{ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\) hay tam giác cân
Cho a,b,c thỏa mãn :
\(\dfrac{1}{a+b+c}=\dfrac{a+4b-c}{c}=\dfrac{b+4c-a}{a}=\dfrac{c+4a-b}{b}\)
Tính: \(P=\left(2+\dfrac{a}{b}\right)\left(3+\dfrac{b}{c}\right)\left(4+\dfrac{c}{a}\right)\)
Ai giải giúp mik với mik đag cần
Cho a, b, c là độ dài 3 cạnh của 1 tam giác. CMR: \(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)