cho a,b,c >0
chứng minh rằng
\(\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{a^2+c^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\ge\dfrac{a+b+c}{\sqrt{2}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 3 2022
\(\sum\dfrac{a}{\sqrt{ab+b^2}}=\sum\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\sum\dfrac{2\sqrt{2}a}{2b+a+b}=2\sqrt{2}\sum\dfrac{a}{a+3b}\)
\(=2\sqrt{2}\sum\dfrac{a^2}{a^2+3ab}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\)
\(=\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3\sqrt{2}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2022
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2021
\(\dfrac{P}{\sqrt{2}}=\dfrac{a}{\sqrt{2b\left(a+b\right)}}+\dfrac{b}{\sqrt{2c\left(b+c\right)}}+\dfrac{c}{\sqrt{2a\left(a+c\right)}}\)
\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2a}{2b+a+b}+\dfrac{2b}{2c+b+c}+\dfrac{2c}{2a+a+c}\)
\(\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)=2\left(\dfrac{a^2}{a^2+3ab}+\dfrac{b^2}{b^2+3bc}+\dfrac{c^2}{c^2+3ca}\right)\)
\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)
\(\Rightarrow P\ge\dfrac{3\sqrt{2}}{2}\) (đpcm)
8 tháng 8 2021
\(\dfrac{a}{\sqrt{ab+b^2}}=\dfrac{\sqrt{2}.a}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{\sqrt{2}.a}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)
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
4 tháng 2 2018
Ko lq nhưng ta chuẩn hóa \(a+b+c=3\). So:
\(M\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{3}{2}\)
AH
Akai Haruma
Giáo viên
14 tháng 3 2018
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 \(a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 10 2021
\(P\sqrt{2}\ge\dfrac{a^2}{\sqrt{b^2+c^2}}+\dfrac{b^2}{\sqrt{c^2+a^2}}+\dfrac{c^2}{\sqrt{a^2+b^2}}\)
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
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow VT=\dfrac{y^2+z^2-x^2}{2x}+\dfrac{x^2+z^2-y^2}{2y}+\dfrac{x^2+y^2-z^2}{2z}\)
\(VT\ge\dfrac{\left(y+z\right)^2}{4x}+\dfrac{\left(x+z\right)^2}{4y}+\dfrac{\left(x+y\right)^2}{4z}-\dfrac{1}{2}\left(x+y+z\right)\)
\(VT\ge\dfrac{\left(2x+2y+2z\right)^2}{4\left(x+y+z\right)}-\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\left(x+y+z\right)\)
\(VT\ge\dfrac{1}{2}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\)
\(VT\ge\dfrac{1}{2}\left(\sqrt{\dfrac{1}{2}\left(a+b\right)^2}+\sqrt{\dfrac{1}{2}\left(b+c\right)^2}+\sqrt{\dfrac{1}{2}\left(c+a\right)^2}\right)\)
\(VT\ge\dfrac{a+b+c}{\sqrt{2}}\) (đpcm)