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3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
Áp dụng BĐT Cô si Ta có : \(\dfrac{a}{b^2+1}=a-\dfrac{ab^2}{b^2+1}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)
\(\dfrac{b}{c^2+1}=b-\dfrac{c^2b}{c^2+1}\ge b-\dfrac{c^2b}{2c}=b-\dfrac{cb}{2}\)
\(\dfrac{c}{a^2+1}=c-\dfrac{a^2c}{a^2+1}\ge c-\dfrac{a^2c}{2a}=c-\dfrac{ac}{2}\)
Cộng ba vế BĐT lại ta được:
\(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge a+b+c-\left(\dfrac{ab+bc+ac}{2}\right)\)
Ta có đánh giá quen thuộc \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{9}{3}=3\)
\(\Rightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)(ĐPCM)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
Ta có : \(\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\le\dfrac{a^3+b^3+c^3}{2abc}+3\)
\(\Leftrightarrow\dfrac{a^2+b^2+c^2}{a^2+b^2}+\dfrac{a^2+b^2+c^2}{b^2+c^2}+\dfrac{a^2+b^2+c^2}{c^2+a^2}\le\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{b^3}{2abc}+3\)( vì \(a^2+b^2+c^2=1\) )
\(\Leftrightarrow3+\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\le\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}+3\)
\(\Leftrightarrow\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\le\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
Mà theo bất đẳng thức cô-si , ta có : \(b^2+c^2\ge2bc\)\(\Rightarrow\dfrac{a^2}{b^2+c^2}\le\dfrac{a^2}{2bc}\)
Tương tự ta cũng có : \(\dfrac{b^2}{c^2+a^2}\le\dfrac{b^2}{2ca},\dfrac{c^2}{a^2+b^2}\le\dfrac{c^2}{2ab}\)
Cộng các bất đẳng thức trên lại với nhau ta được :
\(\Leftrightarrow\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\le\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
Do đó bất đẳng thức ban đầu được chứng minh .
Lời giải:
Áp dụng BĐT AM-GM cho các số dương:
\(a^2+bc\geq 2\sqrt{a^2bc}; b^2+ac\geq 2\sqrt{b^2ac}; c^2+ab\geq 2\sqrt{c^2ab}\)
Do đó:
\(\text{VT}=\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2bc}}+\frac{1}{2\sqrt{b^2ac}}+\frac{1}{2\sqrt{c^2ab}}\)
hay \(\text{VT}\leq \frac{\sqrt{bc}+\sqrt{ac}+\sqrt{ab}}{2abc}(*)\)
Tiếp tục áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} \sqrt{bc}\leq \frac{b+c}{2}\\ \sqrt{ac}\leq \frac{a+c}{2}\\ \sqrt{ab}\leq \frac{a+b}{2}\end{matrix}\right.\Rightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq a+b+c(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\leq \frac{a+b+c}{2abc}\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
Áp dụng BĐT AM - GM, ta có:
\(a^2+2b^2+3\)
\(=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)
\(\ge2ab+2b+2\)
Tương tự, ta có: \(b^2+2c^2+3\ge2bc+2c+2\) và \(c^2+2a^2+3\ge2ac+2a+2\)
\(VT=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)
\(\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ac+2a+2}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ac+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{abc}{bc+c+abc}+\dfrac{abc}{ac+a^2bc+abc}\right)\) (Thay abc = 1)
\(=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)\)
\(=\dfrac{1}{2}\times\dfrac{1+ab+b}{ab+b+1}\)
\(=\dfrac{1}{2}=VP\left(\text{đ}pcm\right)\)
Dấu "=" xảy ra khi a = b = c = 1
áp dụng cô si ta có :
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\ge\dfrac{\left(1+1+1\right)^2}{2a+b+2b+c+2c+a}\)
\(=\dfrac{9}{3\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)