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bài 1)
ta có \(\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)
\(\Rightarrow a^2-2ab+b^2+a^2-2a+1+b^2-2b+1\ge0\)
=> \(a^2+b^2+1\ge ab+a+b\)
a/ Cho x, y ≥ 1. Chứng minh: 1/(1 + x^2) + 1/(1 + y^2) ≥ 2/(1 + xy)
b/ Đề:...Tìm GTLN
Có:
\(\dfrac{1}{4x^2-4x+2}=\dfrac{1}{\left(2x-1\right)^2+1}\le\dfrac{1}{2}\forall x\ge1\)
\(\dfrac{1}{9y^2+6y+2}=\dfrac{1}{\left(3y+1\right)^2+1}\le\dfrac{1}{2}\forall y\ge0\)
\(\Rightarrow A=\dfrac{1}{4x^2-4x+2}+\dfrac{1}{9y^2+6y+2}\le\dfrac{1}{2}+\dfrac{1}{2}=1\)
Vậy MAXA = 1 khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira,
Nguyễn Thị Ngọc Thơ, @tth_new
help me! cần gấp lắm ạ!
thanks nhiều!
b/ \(a-\frac{1}{a}=\sqrt{a}+\frac{1}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}-\frac{1}{\sqrt{a}}=1\)
\(\Leftrightarrow a+\frac{1}{a}-2=1\)
\(\Leftrightarrow a+\frac{1}{a}=3\)
\(\Leftrightarrow a^2+\frac{1}{a^2}+2=9\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2=5\)
\(\Leftrightarrow a-\frac{1}{a}=\sqrt{5}\)
a/ Ta có: \(x=\frac{1-5y}{2}\) thê vô ta được
\(x^2+y^2=y^2+\left(\frac{1-5y}{2}\right)^2=\frac{29y^2-10y+1}{4}\)
\(=\frac{1}{116}\left(29^2y^2-290y+29\right)=\frac{1}{116}\left[\left(29^2y^2-2.29y.5+25\right)+4\right]\)
\(=\frac{1}{116}\left[\left(29y-5\right)^2+4\right]\ge\frac{4}{116}=\frac{1}{29}\)
\(1.\)\(a^3b^3\left(a^2-ab+b^2\right)\le\frac{\left(a+b\right)^8}{256}\)
\(\Leftrightarrow a^3b^3\left(a^2-ab+b^2\right)\left(a+b\right)\le\frac{\left(a+b\right)^9}{256}\)
\(\Leftrightarrow a^3b^3\left(a+b\right)^3\left(a^3+b^3\right)\le\frac{\left(a+b\right)^{12}}{256}\)
\(VT=ab\left(a+b\right).ab\left(a+b\right).ab\left(a+b\right).\left(a^3+b^3\right)\)
\(\le\left(\frac{ab\left(a+b\right)+ab\left(a+b\right)+ab\left(a+b\right)+\left(a^3+b^3\right)}{4}\right)^4\)
\(\le\frac{\left(a^3+3a^2b+3ab^2+b^3\right)^4}{256}\)
\(\le\frac{\left(a+b\right)^{12}}{256}\left(đpcm\right).\)
\(2.\) \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge2\)
\(\Leftrightarrow\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}\)
\(\ge\frac{b}{1+b}+\frac{c}{1+c}\)
\(\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\\\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\end{cases}}\)
\(\Rightarrow\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}\ge8\sqrt{\frac{a^2b^2c^2}{\left(1+a\right)^2.\left(1+b\right)^2.\left(1+c\right)^2}}\)\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow\) \(1\ge8abc\)
\(\Leftrightarrow\) \(abc\ge\frac{1}{8}\left(đpcm\right).\)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
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
a) \(2x+3y=4\Rightarrow x=\frac{4-3y}{2}\)
Lúc đó thì\(2x^2+3y^2=2\left(\frac{4-3y}{2}\right)^2+3y^2=\frac{\left(4-3y\right)^2+6y^2}{2}=\frac{9y^2-24y+16+6y^2}{2}\)\(=\frac{15y^2-24y+16}{2}=\frac{15\left(y^2-\frac{24}{15}+\frac{16}{25}\right)+\frac{32}{5}}{2}=\frac{15\left(y-\frac{4}{5}\right)^2+\frac{32}{5}}{2}\ge\frac{\frac{32}{5}}{2}=\frac{16}{5}\)
Đẳng thức xảy ra khi x = y = 4/5
b) \(3a-5b=8\Rightarrow a=\frac{5b+8}{3}\)
Lúc đó thì \(7a^2+11b^2=7\left(\frac{5b+8}{3}\right)^2+11b^2=\frac{7\left(5b+8\right)^2+99b^2}{9}\)\(=\frac{175b^2+560b+448+99b^2}{9}=\frac{274b^2+560b+448}{9}\)\(=\frac{274\left(b^2+\frac{280}{137}b+\left(\frac{140}{137}\right)^2\right)+\left(448-274.\left(\frac{140}{137}\right)^2\right)}{9}=\frac{274\left(b+\frac{140}{137}\right)^2+\frac{22176}{137}}{9}\ge\frac{2464}{137}\)
Đẳng thức xảy ra khi a = 132/137; b = -140/137