cho a,b dương tỏa mãn: \(\sqrt{a}+\sqrt{b}\ge2\)
CMR: \(\sqrt{a^3}+\sqrt{b^3}\ge a+b\)
K
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22 tháng 8 2017
doan thi khanh linh câm cái mồm đi.đã ngu lại còn thích k
22 tháng 8 2017
áp dụng co si ta có:
\(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\frac{2\sqrt{bc}}{\sqrt{a}}+\frac{2\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}\)
\(=\left(\frac{\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}\right)+\left(\frac{\sqrt{ca}}{\sqrt{b}}+\frac{\sqrt{ab}}{\sqrt{c}}\right)+\left(\frac{\sqrt{ab}}{\sqrt{c}}+\frac{\sqrt{bc}}{\sqrt{a}}\right)\)
\(\ge2\sqrt{a}+2\sqrt{b}+2\sqrt{c}=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
\(\Rightarrow Q.E.D\)
N
8 tháng 9 2019
Ta co:
\(\sqrt{2\left(b+1\right)}\le\frac{b+3}{2}\Rightarrow\frac{a}{\sqrt{2\left(b+1\right)}}\ge\frac{2a}{b+3}\)
Tuong tu:\(\frac{b}{\sqrt{2\left(c+1\right)}}\ge\frac{2b}{c+3};\frac{c}{\sqrt{2\left(a+1\right)}}\ge\frac{2c}{a+3}\)
\(\Rightarrow\frac{1}{\sqrt{2}}\left(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\right)\ge2\left(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\right)\)
\(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\)
\(=\frac{a^2}{ab+3a}+\frac{b^2}{bc+3b}+\frac{c^2}{ca+3c}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca+9}\ge\frac{\left(a+b+c\right)^2}{\frac{\left(a+b+c\right)^2}{3}+9}=\frac{9}{\frac{9}{3}+9}=\frac{3}{4}\)
\(\Rightarrow2\left(\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{a+3}\right)\ge\frac{3}{2}\)
Hay \(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\ge\frac{3\sqrt{2}}{2}\)
Dau '=' xay ra khi \(a=b=c=3\)
28 tháng 8 2019
\(VP^2\le2\left(a+b+c\right)\left(a^2+b^2+c^2\right)\) (1)
\(VT^2=\left(\frac{a^4}{a}+\frac{b^4}{b}+\frac{c^4}{c}\right)^2\ge\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)^3}{\left(a+b+c\right)^2}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^6}{27\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)\left(a+b+c\right)^3}{27}\)
\(\ge\frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)\left(3\sqrt[3]{abc}\right)^3}{27}=2\left(a+b+c\right)\left(a^2+b^2+c^2\right)\ge VP^2\) (2)
Mà VT và VP đều dường nên từ (1) và (2) suy ra đpcm
Dấu "=" xảy ra khi \(a=b=c=\sqrt[3]{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 3 2022
Đặt vế trái BĐT cần chứng minh là P
Ta có:
\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2011}\)
Đồng thời: \(\left\{{}\begin{matrix}y^2+z^2-x^2=2a^2\\z^2+x^2-y^2=2b^2\\x^2+y^2-z^2=2c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z+z+x+x+y\right)^2}{2x+2y+2z}-\left(x+y+z\right)\right)=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)
F
6 tháng 4 2022
Đặt \(x=\sqrt{a};y=\sqrt{b};z=\sqrt{c}\) \(\Rightarrow xyz=1\) (x;y;z > 0 do a;b;c>0)
Cần c/m : \(VT=\dfrac{y^2+z^2}{x}+\dfrac{x^2+z^2}{y}+\dfrac{x^2+y^2}{z}\ge x+y+z+3=VP\)
Dễ dàng c/m : VT \(\ge2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\) (1)
Thấy : \(\dfrac{xy}{z}+\dfrac{xz}{y}\ge2x\) . CMTT : \(\dfrac{xz}{y}+\dfrac{yz}{x}\ge2z;\dfrac{yz}{x}+\dfrac{xy}{z}\ge2y\)
Suy ra : \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge x+y+z\)
Có : \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge3\sqrt[3]{xyz}=3\)
Suy ra : \(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\ge x+y+z+3\left(2\right)\)
Từ (1) ; (2) suy ra : \(VT\ge VP\)
" = " \(\Leftrightarrow x=y=z=1\Leftrightarrow a=b=c=1\)
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Bài 1:
Áp dụng BĐT Bunhiacopxky ta có:
$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$
$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$
$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Bài 2:
Áp dụng BĐT Bunhiacopxky:
$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$
$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$
$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$
$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>
\(\left\{{}\begin{matrix}\sqrt{a}=x\ge0\\\sqrt{b}=y\ge0\end{matrix}\right.\Rightarrow x+y\ge2\)
Ta cm bđt sau: \(2\left(x^3+y^3\right)\ge\left(x+y\right)\left(x^2+y^2\right)\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (đúng)
Mà: \(x+y\ge2\Rightarrow x^3+y^3\ge x^2+y^2\)
\(\Rightarrow\sqrt{a^3}+\sqrt{b^3}\ge a+b\)(đpcm)
\("="\Leftrightarrow a=b=1\)