Cho a,b,c >0; a+b+c=6. Chứng minh rằng
\(\dfrac{a+b}{c^2+4}+\dfrac{b+c}{a^2+4}+\dfrac{c+a}{b^2+4}\ge\dfrac{3}{2}\)
Cho a,b,c > 0
Chứng minh rằng: \(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}+\dfrac{c^4}{a^2}\)
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2\ge5\sqrt[5]{\dfrac{a^{20}b^2}{b^{12}}}=5.\dfrac{a^4}{b^2}\)
\(\Rightarrow4.\dfrac{a^5}{b^3}+b^2\ge5.\dfrac{a^4}{b^2}\)
Tương tự: \(4.\dfrac{b^5}{c^3}+c^2\ge5\dfrac{b^4}{c^2};4\dfrac{c^5}{a^3}+a^2\ge5.\dfrac{c^4}{a^2}\)
\(\Rightarrow4\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)
Lại có: \(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5a^2\)
\(\Rightarrow2.\dfrac{a^5}{b^3}+3b^2\ge5a^2\), tương tự: \(2.\dfrac{b^5}{c^3}+3c^2\ge5b^2;2\dfrac{c^5}{a^3}+3a^2\ge5c^2\)
\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}+4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5.\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)
\(\Rightarrow dpcm\)
giả sử \(a>b>c>0\) thì ta có :
\(\dfrac{a^4}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^4}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^4}{a^2}\left(\dfrac{c}{a}-1\right)\ge\dfrac{2a^2b}{c}+\dfrac{c^5}{a^3}-\dfrac{c^4}{a^2}\)
\(\ge\dfrac{2c^4b}{a}-\dfrac{c^4}{a^2}=\dfrac{c^4}{a}\left(2b-\dfrac{1}{a}\right)>0\)
làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
mấy câu cậu câu đăng khác bn làm tương tự nha . nếu bn lm không được thì có j mk lm luôn cho còn h mk bạn rồi :(
Cho \(a,b,c>0\) thỏa mãn \(a^4+b^4+c^4=3\). Chứng minh:
\(\dfrac{a^2}{b^3+1}+\dfrac{b^2}{c^3+1}+\dfrac{c^2}{a^3+1}\ge\dfrac{3}{2}\)
1. Cho a,b,c t/m: \(\left\{{}\begin{matrix}a\ge\dfrac{4}{3}\\b\ge\dfrac{4}{3}\\c\ge\dfrac{4}{3}\end{matrix}\right.\) và \(a+b+c=6\)
\(CMR:\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\ge\dfrac{6}{5}\)
2. Cho x,y >0 t/m: \(2x+3y-13\ge0\)
Tìm min \(P=x^2+3x+\dfrac{4}{x}+y^2+\dfrac{9}{y}\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
Cho a,b,c > 0 thỏa mãn \(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}=3\). Chứng minh rằng:
\(N=\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}+\dfrac{c^4}{a^2}\ge3\)
Áp dụng \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)
\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a,b,c >0 Chứng minh rằng:
a) \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)
b) \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
1. Chứng minh: \(a^6+b^6+c^6\ge a^5b+ac^5+b^5c\) với \(a,b,c\ge0\)
2. Chứng minh rằng: với a,b,c > 0 thì \(\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{a^2+c^2}+\dfrac{c^2}{a^2+b^2}\ge\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
3. Chứng minh rằng: \(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3\) với a,b,c > 0.
4. Cho a,b,c là độ dài 3 cạnh của tam giác. Chứng minh: \(\dfrac{1}{a+b};\dfrac{1}{a+c};\dfrac{1}{b+c}\) là độ dài của tam giác.
3) Biến đổi tương đương:
\(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3\) (1)
\(\Leftrightarrow\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+6\left(a^3+c^3+b^3\right)\)
\(\ge\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)
\(\Leftrightarrow\left[a^3+b^3-ab\left(a+b\right)\right]+\left[a^3+c^3-ac\left(a+c\right)\right]+\left[b^3+c^3-bc\left(b+c\right)\right]\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(a+c\right)\left(a-c\right)^2+\left(b+c\right)\left(b-c\right)^2\ge0\) luôn đúng do a, b, c > 0
=> (1) đúng
Dấu "=" xảy ra khi a = b = c
4) Ta có: a+b>c ; b+c>a; a+c>b
Xét \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
Tương tự: \(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy suy ra được điều phải chứng minh
2) Xét: \(\dfrac{a^2}{b^2+c^2}-\dfrac{a}{b+c}=\dfrac{a\left(ab+ac-b^2-c^2\right)}{\left(b^2+c^2\right)\left(b+c\right)}=\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\left(1\right)\)
Tương tự:
\(\dfrac{b^2}{c^2+a^2}-\dfrac{b}{c+a}=\dfrac{bc\left(b-c\right)+ba\left(b-a\right)}{\left(c^2+a^2\right)\left(c+a\right)}\left(2\right)\)
\(\dfrac{c^2}{a^2+b^2}-\dfrac{c}{a+b}=\dfrac{ca\left(c-a\right)+cb\left(c-b\right)}{\left(c^2+a^2\right)\left(c+a\right)}\left(3\right)\)
Cộng (1),(2),(3) vế theo vế ta được:
\(\left(\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\right)-\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\dfrac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}\right]+ac\left(a-c\right)\left[\dfrac{1}{b^2+c^2\left(b+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]+bc\left(b-c\right)\left[\dfrac{1}{\left(a^2+c^2\right)\left(a+c\right)}-\dfrac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
Giả sử \(a\ge b\ge c>0\) thì các biểu thức trong ngoặc tròn, vuông không âm
=> đpcm
Cho a,b,c>0 .
Chứng minh rằng \(\dfrac{a^4}{a^3+b^3^{ }}+\dfrac{b^4}{b^3+c^3}+\dfrac{c^4}{c^3+a^3}\)≥\(\dfrac{a+b+c}{2}\)
Bài tập 1:
Cho x,y > 0. Chứng minh rằng: ( 3x+3y )(\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\)) ≥4
Bài tập 2: Cho a,b,c> 0. Chứng minh rằng:
a) \(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\)≤\(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
b) \(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\)≤\(\dfrac{3}{2}\)
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
ta có:
\(1+a^2\ge2\sqrt{a^2.1}=2a\Rightarrow\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
tương tự, ta có: \(\dfrac{b}{b^2+1}\le\dfrac{1}{2}\); \(\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
công từ vế với nhau, ta có điều cần phải chứng minh
Bài 1 : Cho a, b, c > 0. Chứng minh rằng : \(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}>=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Bài 2 : Cho a, b, c là các số thực. Chứng minh rằng : a4+b4+c4 >= abc(a+b+c)
Bài 2:
Áp dụng BĐT: \(x^2+y^2+z^2\ge xy+yz+xz\), ta có:
\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2\) (1)
Lại áp dụng tương tự ta có:
\(\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\ge ab^2c+abc^2+a^2bc\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\) (2)
Từ (1) và (2) suy ra:
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Bài 1:
Áp dụng BĐT Cô -si, ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\sqrt[3]{\dfrac{b^2}{c^3}.\dfrac{1}{b}.\dfrac{1}{b}}=\dfrac{3}{c}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng vế theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
p/s: không chắc lắm, có gì sai xót xin giúp đỡ