Cho \(a,b,c\) là ba số dương. Chứng minh rằng
\(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\ge\dfrac{36}{3a+2b+c}\).
cho 3 số thực dương a,b,c thỏa mãn (3a+2b)(3a+2c)=16bc. Chứng minh rằng
a) b+c ≥ 3a
b)\(\dfrac{a}{b+c}+\dfrac{b+c}{a}\) ≥ \(\dfrac{10}{3}\)
Lời giải:
\((3a+2b)(3a+2c)=16bc\)
\(\Leftrightarrow 9a^2+6a(b+c)=12bc\)
Theo BĐT Cô-si \(4bc\leq (b+c)^2\Rightarrow 9a^2+6a(b+c)\leq 3(b+c)^2\)
\(\Rightarrow 3a^2+2a(b+c)\leq (b+c)^2\)
\(\Leftrightarrow (b+c)^2-3a^2-2a(b+c)\geq 0\)
\(\Leftrightarrow (b+c)^2-9a^2-2a(b+c)+6a^2\geq 0\)
\(\Leftrightarrow (b+c-3a)(b+c+3a)-2a(b+c-3a)\geq 0\)
\(\Leftrightarrow (b+c-3a)(b+c+a)\geq 0\)
Vì $a+b+c>0$ nên \(b+c-3a\geq 0\Rightarrow b+c\geq 3a\) (đpcm)
b) Áp dụng BĐT Cô-si và kết quả phần a:
\(\frac{a}{b+c}+\frac{b+c}{a}=\frac{a}{b+c}+\frac{b+c}{9a}+\frac{8(b+c)}{9a}\)
\(\geq 2\sqrt{\frac{a}{b+c}.\frac{b+c}{9a}}+\frac{8(b+c)}{9a}=\frac{2}{3}+\frac{8(b+c)}{9a}\geq \frac{2}{3}+\frac{8.3a}{9a}=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)
Ta có đpcm.
Cho a,b,c là các số dương, chứng minh rằng
\(\dfrac{2a^2}{2b+c}+\dfrac{2b^2}{2a+c}+\dfrac{c^2}{4a+4b}\ge\dfrac{1}{4}\left(2a+2b+c\right)\)
\(P=\dfrac{4a^2}{4b+2c}+\dfrac{4b^2}{4a+2c}+\dfrac{c^2}{4a+4b}\ge\dfrac{\left(2a+2b+c\right)^2}{8a+8b+4c}\)
\(=\dfrac{\left(2a+2b+c\right)^2}{4\left(2a+2b+c\right)}=\dfrac{1}{4}\left(2a+2b+c\right)\)
Cho a,b,c là các số thực dương. Chứng minh rằng:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
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 \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)
Cho a, b, c là các số dương thỏa mãn: \(a^2+2b^2\le3c^2\). Chứng minh: \(\dfrac{1}{a}+\dfrac{2}{b}\ge\dfrac{3}{c}\)
Tham khảo: https://lazi.vn/edu/exercise/cho-a-b-c-la-cac-so-duong-thoa-man-a2-2b2-3c2-chung-minh-1-a-2-b-3-c
Cho a , b , c là các số thực dương . Chứng minh rằng
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Cho các số thực dương a,b. Chứng minh rằng:
a/ \(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{9ab}{a^2+b^2}\ge\dfrac{13}{2}\)
b/ \(\dfrac{a}{3b}+\dfrac{b\left(a+b\right)}{a^2+ab+b^2}\ge1\)
c/ \(\dfrac{a}{2b}+\dfrac{2b}{a+b}+\dfrac{ab}{2\left(a^3+2b^3\right)}\ge\dfrac{5}{3}\)
a) Sai với \(a=1,b=2\)
b)
Thực hiện biến đổi tương đương:
\(\frac{a}{3b}+\frac{b(a+b)}{a^2+ab+b^2}\geq 1\)
\(\Leftrightarrow \frac{a}{3b}+\frac{b(a+b)+a^2}{a^2+ab+b^2}-\frac{a^2}{a^2+ab+b^2}\geq 1\)
\(\Leftrightarrow \frac{a}{3b}-\frac{a^2}{a^2+ab+b^2}\geq 0\)
\(\Leftrightarrow \frac{1}{3b}-\frac{a}{a^2+ab+b^2}\geq 0\)
\(\Leftrightarrow \frac{a^2+ab+b^2-3ab}{3b(a^2+ab+b^2)}\geq 0\)
\(\Leftrightarrow \frac{(a-b)^2}{3b(a^2+ab+b^2)}\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b$
c) BĐT sai với \(a=1,b=2\)
cho ba số dương a,b,c .Chứng minh rằng \(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(a+c\right)}+\dfrac{1}{c^2\left(b+a\right)}\ge\dfrac{3}{2}\)
Nhìn qua đã biết là đề sai rồi bạn
Cho \(a,b,c\) các giá trị lớn ví dụ \(a=b=c=2\) là thấy sai ngay
Cho ba số dương a,b,c sao cho : abc=1 .Chứng minh :
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{3}{2}\left(a+b+c-1\right)\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)
BĐT trở thành:
\(\dfrac{y^2}{xz}+\dfrac{z^2}{xy}+\dfrac{x^2}{yz}\ge\dfrac{3}{2}\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}-1\right)\)
\(\Leftrightarrow2\left(x^3+y^3+z^3\right)+3xyz\ge3x^2y+3y^2z+3z^2x\)
Áp dụng BĐT Schur ta có:
\(x^3+y^3+z^3+3xyz\ge x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\)
\(\Rightarrow VT\ge\left(x^3+xy^2\right)+\left(y^3+yz^2\right)+\left(z^3+zx^2\right)+x^2y+y^2z+z^2x\ge3\left(x^2y+y^2z+z^2x\right)\)