PH
1
K
Khách
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30 tháng 9 2017
\(A=\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)
\(=\frac{a^3}{a^2+ab+bc+ca}+\frac{b^3}{b^2+ab+bc+ca}+\frac{c^3}{c^2+ab+bc+ca}\)
\(=\frac{a^3}{\left(a+b\right)\left(c+a\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(c+a\right)\left(b+c\right)}\)
đến đây áp dụng cô si 3 số là đc
7 tháng 5 2019
Ta có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow a^2+b^2+c^2\ge2\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(a^4+b^4+c^4\ge\frac{\left(a^2+b^2+c^2\right)^2}{3}\ge\frac{4}{3}\)
\(\Rightarrow a^4+b^4+c^4\ge\frac{4}{3}\left(đpcm\right)\)
Dấu '=' xảy ra khi\(\hept{\begin{cases}a=b=c\\ab+bc+ca=2\end{cases}\Leftrightarrow a=b=c=\sqrt{\frac{2}{3}}}\)
4 tháng 2 2021
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NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 2 2020
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)
Cộng vế với vế ta có đpcm
HB
1
UK
26 tháng 2 2018
Bạn post nhiều bài BĐT hay thật
Đặt \(\left(a;b;c\right)=\left(\dfrac{2x}{y+z};\dfrac{2y}{z+x};\dfrac{2z}{x+y}\right)\)
BĐT trở thành:
\(\sum_{cyc}\dfrac{x}{y+z}\ge\sum_{cyc}\dfrac{2xy}{\left(x+y\right)\left(x+z\right)}\)
Sử dụng AM-GM, ta có:
\(VP\le\sum_{cyc}xy\left[\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x+z\right)^2}\right]=\sum_{cyc}\dfrac{xy}{\left(z+x\right)^2}+\sum_{cyc}\dfrac{xy}{\left(y+z\right)^2}=\sum_{cyc}\dfrac{xy}{\left(y+z\right)^2}+\sum\dfrac{zx}{\left(y+z\right)^2}=\sum_{cyc}\dfrac{x}{y+z}=VT\)
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