Cho \(a,b,c>0\) thỏa mãn \(a^4+b^4+c^4=3\). Chứng minh:
\(\dfrac{a^2}{b^3+1}+\dfrac{b^2}{c^3+1}+\dfrac{c^2}{a^3+1}\ge\dfrac{3}{2}\)
Cho ba số a,b,c thỏa mãn a+b+c=\(\dfrac{3}{2}\). Chứng minh:
\(a^2+b^2+c^2\ge\dfrac{3}{4}\)
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 $a=b=c=\frac{1}{2}$.
a,b,c>0 thỏa mãn `a^4 +b^4 +c^4 =3`. CMR \(\dfrac{a^2}{b^3+1}+\dfrac{b^2}{c^3+1}+\dfrac{c^2}{a^3+1}>=\dfrac{3}{2}\)
a,b,c>0 thỏa mãn `a^4 +b^4 +c^4 =3`. CMR: \(\dfrac{a^2}{b^3+1}+\dfrac{b^2}{c^3+1}+\dfrac{c^2}{a^3+1}>=\dfrac{3}{2}\)
B1: Cho \(0\le a,b,c\le2\) thỏa mãn \(a+b+c=3\). CMR: \(a^2+b^2+c^2\le5\)
B2: Cho \(a,b\ge0\) thỏa mãn \(a^2+b^2=a+b\). TÌm GTLN \(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}\)
B3: CMR: \(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
Cho \(a;b;c\ge\dfrac{4}{3}\) thỏa mãn \(a+b+c=6\)
Tìm min: \(A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\)
Lời giải:
Xét:
$\frac{a}{a^2+1}-\left(\frac{16}{25}-\frac{3}{25}a\right)=\frac{(a-2)^2(3a-4)}{25(a^2+1)}\geq 0$ với mọi $a\geq \frac{4}{3}$
$\Rightarrow \frac{a}{a^2+1}\geq \frac{16}{25}-\frac{3}{25}a$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế, suy ra:
$A\geq \frac{48}{25}-\frac{3}{25}(a+b+c)=\frac{6}{5}$
Vậy $A_{\min}=\frac{6}{5}$.
Giá trị này đạt tại $a=b=c=2$
Cho 3 số dương a, b, c thỏa mãn: ab+bc+ca=3. Chứng minh: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\)
\(ab+bc+ac=3\)
Ta có:
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) ( đúng với mọi \(ab\ge1\))
Giả sử:\(ab\ge1\)
\(\Rightarrow\dfrac{2}{ab+1}+\dfrac{1}{c^2+1}\ge\dfrac{2c^2+2+ab+1}{\left(ab+1\right)\left(c^2+1\right)}=\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\)
Giả sử: \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\)(đúng)
\(\Leftrightarrow2\left(2c^2+ab+3\right)\ge3\left(ab+1\right)\left(c^2+1\right)\)
\(\Leftrightarrow4c^2+2ab+6\ge3\left(abc^2+ab+c^2+1\right)\)
\(\Leftrightarrow4c^2+2ab+6\ge3abc^2+3ab+3c^2+3\)
\(\Leftrightarrow c^2-ab-3abc^2+3\ge0\)
\(\Leftrightarrow c^2-ab-3abc^2+ab+ac+bc\ge0\) ( vì \(ab+ac+bc=3\) )
\(\Leftrightarrow c^2+ac+bc-3abc^2\ge0\)
\(\Leftrightarrow c+a+b-3abc\ge0\)
\(\Leftrightarrow c+a+b\ge3abc\)
Ta có:
\(3\left(c+a+b\right)=\left(ab+ac+bc\right)\left(c+a+b\right)\) ( vì \(ab+ac+bc=3\) )
Áp dụng BĐT AM-GM, ta có:
\(\left(ab+ac+bc\right)\left(c+a+b\right)\ge9abc\)
\(\Rightarrow a+b+c\ge3abc\)
\(\Rightarrow\) \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\) ( luôn đúng )
\(\Rightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\) ( đfcm )
Dấu "=" xảy ra khi \(a=b=c=1\)
Hình như sai đề rồi bạn ạ, dấu ≥ phải là ≤
cho a,b,c >0 thỏa mãn \(a^2+b^2+c^2=3\) chứng minh rằng \(\dfrac{a}{ab+3}+\dfrac{b}{bc+3}+\dfrac{c}{ca+3}\le\dfrac{3}{4}\)
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 là các số thực dương thỏa mãn abc=1.Chứng minh rằng \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\ge\dfrac{1}{2}\)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\Rightarrow xyz=1\)
Đặt vế trái BĐT cần chứng minh là P, ta có:
\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)
\(P=\dfrac{1}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}+\dfrac{1}{\left(y^2+z^2\right)+\left(z^2+1\right)+2}+\dfrac{1}{\left(z^2+x^2\right)+\left(x^2+1\right)+2}\)
\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)
\(P\le\dfrac{1}{2}\left(\dfrac{xz}{xz\left(xy+y+1\right)}+\dfrac{x}{x\left(yz+z+1\right)}+\dfrac{1}{zx+x+1}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x.xyz+xyz+xz}+\dfrac{x}{xyz+xz+1}+\dfrac{1}{xz+x+1}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x+1+xz}+\dfrac{x}{1+xz+1}+\dfrac{1}{xz+x+1}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)