cho a,b,c >0
chứng minh rằng
\(\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{a^2+c^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\ge\dfrac{a+b+c}{\sqrt{2}}\)
Cho 3 số dương a;b;c thoả mãn : \(\sqrt{a^2+b^2}\text{+}\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\text{=}\sqrt{2011}\)
Chứng minh rằng : \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Cho 3 số dương a, b, c thoả mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\). Chứng minh rằng: \(\dfrac{a^2}{b+c}+\dfrac{a^2}{b+c}+\dfrac{a^2}{b+c}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
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) \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)
Cho a,b,c là 3 số thực dương tùy ý Chứn minh rằng
\(\dfrac{a}{\sqrt{ab+b^2}}+\dfrac{b}{\sqrt{bc+b^2}}+\dfrac{c}{\sqrt{ac+a^2}}\ge\dfrac{3\sqrt{2}}{2}\)
\(\sum\dfrac{a}{\sqrt{ab+b^2}}=\sum\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\sum\dfrac{2\sqrt{2}a}{2b+a+b}=2\sqrt{2}\sum\dfrac{a}{a+3b}\)
\(=2\sqrt{2}\sum\dfrac{a^2}{a^2+3ab}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\)
\(=\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3\sqrt{2}}{2}\)
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 \(a=b=c=1\)
cho a,b,c>0 chứng minh
\(P=\dfrac{a}{\sqrt{ab+b^2}}+\dfrac{b}{\sqrt{bc+c^2}}+\dfrac{c}{\sqrt{ca+a^2}}\ge\dfrac{3\sqrt{2}}{2}\)
\(\dfrac{P}{\sqrt{2}}=\dfrac{a}{\sqrt{2b\left(a+b\right)}}+\dfrac{b}{\sqrt{2c\left(b+c\right)}}+\dfrac{c}{\sqrt{2a\left(a+c\right)}}\)
\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2a}{2b+a+b}+\dfrac{2b}{2c+b+c}+\dfrac{2c}{2a+a+c}\)
\(\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)=2\left(\dfrac{a^2}{a^2+3ab}+\dfrac{b^2}{b^2+3bc}+\dfrac{c^2}{c^2+3ca}\right)\)
\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)
\(\Rightarrow P\ge\dfrac{3\sqrt{2}}{2}\) (đpcm)
\(\dfrac{a}{\sqrt{ab+b^2}}=\dfrac{\sqrt{2}.a}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{\sqrt{2}.a}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)
làm tương tự với \(\dfrac{b}{\sqrt{bc+c^2}};\dfrac{c}{\sqrt{ca+a^2}}\)
\(=>P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\)
\(=2\sqrt{2}\left(\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)
\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+\dfrac{4}{3}\left(ab+bc+ca\right)+\dfrac{8}{3}\left(ab+bc+ca\right)}\right]\)
\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{\dfrac{4}{3}\left(a+b+c\right)^2}\right]=\dfrac{2\sqrt{2}.3}{4}=\dfrac{3\sqrt{2}}{2}\)
dấu"=" xảy ra<=>a=b=c
Cho a,b,c>0 và thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\dfrac{7-abc}{\sqrt{2}}\)
Chứng minh rằng \(M=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{3}{2}\)
Ko lq nhưng ta chuẩn hóa \(a+b+c=3\). So:
\(M\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{3}{2}\)
Cho các số dương a, b, c chứng minh rằng:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)
Lời giải:
Ta có: \(\text{VT}=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=\frac{a^2}{b}-a+b+\frac{b^2}{c}-b+c+\frac{c^2}{a}-c+a\)
\(=\frac{a^2-ab+b^2}{b}+\frac{b^2-bc+c^2}{c}+\frac{c^2-ca+a^2}{a}\)
Áp dụng BĐT AM-GM:
\(\frac{a^2-ab+b^2}{b}+b\geq 2\sqrt{a^2-ab+b^2}\)
\(\frac{b^2-bc+c^2}{c}+c\geq 2\sqrt{b^2-bc+c^2}\)
\(\frac{c^2-ca+a^2}{a}+a\geq 2\sqrt{c^2-ca+a^2}\)
Cộng theo vế:
\(\Rightarrow \text{VT}+(a+b+c)\geq 2(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2})(1)\)
Lại có:
\(\sqrt{a^2-ab+b^2}=\sqrt{\frac{3}{4}(a-b)^2+\frac{1}{4}(a+b)^2}\geq \sqrt{\frac{1}{4}(a+b)^2}=\frac{a+b}{2}\)
TT: \(\sqrt{b^2-bc+c^2}\geq \frac{b+c}{2}; \sqrt{c^2-ca+a^2}\geq \frac{c+a}{2}\)
Suy ra: \(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\geq a+b+c(2)\)
Từ \((1);(2)\Rightarrow \text{VT}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c\)
cho a,b,c dương thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\).
CMR: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\sqrt{2011}}{2}\)
Đặt \(\left(x,y,z\right)=\left(\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+a^2}\right)\).
Ta có \(x+y+z=\sqrt{2011}\).
BĐT cần cm trở thành:
\(\dfrac{y^2+z^2-x^2}{2\sqrt{2}x}+\dfrac{z^2+x^2-y^2}{2\sqrt{2}y}+\dfrac{x^2+y^2-z^2}{2\sqrt{2}z}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
\(\Leftrightarrow\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\ge x+y+z\)
\(\Leftrightarrow\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{y}\right)+\left(\dfrac{y^2}{x}+\dfrac{z^2}{y}+\dfrac{x^2}{z}\right)\ge2\left(x+y+z\right)\).
Theo bđt AM - GM:
\(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{y}=\left(\dfrac{x^2}{y}+y\right)+\left(\dfrac{y^2}{z}+z\right)+\left(\dfrac{z^2}{x}+x\right)-x-y-z\ge2x+2y+2z-x-y-z=x+y+z\).
Tương tự, \(\dfrac{y^2}{x}+\dfrac{z^2}{y}+\dfrac{x^2}{z}\ge x+y+z\).
Dễ có điều phải chứng minh.
\(P\sqrt{2}\ge\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{c^2+a^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2011}\\a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow P2\sqrt{2}\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)
\(P4\sqrt{2}\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)
\(P2\sqrt{2}\ge\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}-\left(x+y+z\right)=x+y+z=\sqrt{2011}\)
\(\Rightarrow P\ge\dfrac{\sqrt{2011}}{2\sqrt{2}}\)
Đề sai