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\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)
\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)
Tương tự và cộng lại:
\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)
\(a+\dfrac{1}{a+1}=\dfrac{a^2+a+1}{a+1}=\dfrac{4a^2+4a+4}{4\left(a+1\right)}=\dfrac{3\left(a+1\right)^2+\left(a-1\right)^2}{4\left(a+1\right)}\ge\dfrac{3\left(a+1\right)^2}{4\left(a+1\right)}=\dfrac{3}{4}\left(a+1\right)\ge\dfrac{3}{2}\sqrt{a}\)
Tương tự: \(b+\dfrac{1}{b+1}\ge\dfrac{3}{2}\sqrt{b}\) ; \(c+\dfrac{1}{c+1}\ge\dfrac{3}{2}\sqrt{c}\)
Nhân vế:
\(VT\ge\dfrac{27}{8}\sqrt{abc}\ge\dfrac{27}{8}\) (đpcm)
Tham khảo:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
\(\dfrac{a^5}{b^3+c^2}+\dfrac{b^3+c^2}{4}+\dfrac{a^4}{2}\ge3\sqrt[3]{\dfrac{a^9.\left(b^3+c^2\right)}{8\left(b^3+c^2\right)}}=\dfrac{3a^3}{2}\)
Tương tự và cộng lại:
\(\Rightarrow M-\dfrac{a^4+b^4+c^4}{2}+\dfrac{a^3+b^3+c^3}{4}+\dfrac{a^2+b^2+c^2}{4}\ge\dfrac{3}{2}\left(a^3+b^3+c^3\right)\)
\(\Rightarrow M\ge\dfrac{a^4+b^4+c^4}{2}+\dfrac{5}{4}\left(a^3+b^3+c^3\right)-\dfrac{3}{4}\)
Mặt khác ta có:
\(\dfrac{1}{2}\left(a^4+b^4+c^4\right)\ge\dfrac{1}{6}\left(a^2+b^2+c^2\right)^2=\dfrac{3}{2}\)
\(\left(a^3+a^3+1\right)+\left(b^3+b^3+1\right)+\left(c^3+c^3+1\right)\ge3\left(a^2+b^2+c^2\right)=9\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge9\Rightarrow a^3+b^3+c^3\ge3\)
\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{15}{4}-\dfrac{3}{4}=...\)
a) Gọi q là công sai của cấp số nhân. Ta có: \(a;b=aq;c=aq^2\).
\(a^2b^2c^2\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{b^2c^2}{a}+\dfrac{a^2c^2}{b}+\dfrac{a^2b^2}{c}\)
\(=\dfrac{\left(a.q\right)^2\left(a.q^2\right)^2}{a}+\dfrac{a^2\left(aq^2\right)^2}{aq}+\dfrac{a^2\left(aq\right)^2}{aq^2}\)
\(=\dfrac{a^2q^2a^2q^4}{a}+\dfrac{a^2a^2q^4}{aq}+\dfrac{a^2a^2q^2}{aq^2}\)
\(=a^3q^6+a^3q^3+a^3\)
\(=\left(a^2q\right)^3+\left(aq\right)^3+a^3\)
\(=c^3+b^3+a^3=a^3+b^3+c^3\).
b) Gọi q là công bội của của cấp số nhân.
Ta có: \(a;b=aq;c=aq^2;d=aq^3\).
\(\left(ab+bc+cd\right)^2=\left(a.aq+aq.aq^2+aq^2.aq^3\right)^2\)
\(=\left(a^2q+a^2q^3+a^2q^5\right)^2=a^4q^2\left(1+q^2+q^4\right)^2\). (1)
\(\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)\)\(=\left(a^2+a^2q^2+a^2q^4\right)\left(a^2q^2+a^2q^4+a^2q^6\right)\)
\(=a^2\left(1+q^2+q^4\right)a^2q^2\left(1+q^2+q^4\right)\)
\(=a^4q^2\left(1+q^2+q^4\right)^2\). (2)
So sánh (1) và (2) ta có điều phải chứng minh.
a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)
c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)
d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)
\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)
\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)
e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)
\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)
\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)
f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)
\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)
\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)
g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)
\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)
\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)
\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)
\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)
\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)
\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)
\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)
\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)
Đặt \(x=\sqrt{\dfrac{a}{bc}}\) ; \(y=\sqrt{\dfrac{b}{ca}}\) ; \(z=\sqrt{\dfrac{c}{ab}}\)
\(\Rightarrow a=\dfrac{1}{yz}\) ; \(b=\dfrac{1}{zx}\) ; \(c=\dfrac{1}{xy}\)
\(\Rightarrow xy+yz+zx=1\)
Khi đó, tồn tại một tam giác ABC sao cho:
\(x=tan\dfrac{A}{2}\) ; \(y=tan\dfrac{B}{2}\) ; \(z=tan\dfrac{C}{2}\)
Thay vào bài toán:
\(A=\dfrac{x^2}{1+x^2}+\sqrt{3}\left(\dfrac{y^2}{1+y^2}+\dfrac{z^2}{1+z^2}\right)\)
\(=\dfrac{tan^2\dfrac{A}{2}}{1+tan^2\dfrac{A}{2}}+\sqrt{3}\left(\dfrac{tan^2\dfrac{B}{2}}{1+tan^2\dfrac{B}{2}}+\dfrac{tan^2\dfrac{C}{2}}{1+tan^2\dfrac{C}{2}}\right)\)
\(=sin^2\dfrac{A}{2}+\sqrt{3}\left(sin^2\dfrac{B}{2}+sin^2\dfrac{C}{2}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}cosA+\dfrac{\sqrt{3}}{2}\left(2-cosB-cosC\right)\)
\(=\dfrac{1+2\sqrt{3}}{2}-\dfrac{1}{2}\left(cosA+\sqrt{3}cosB+\sqrt{3}cosC\right)\)
Xét \(B=cosA+\sqrt{3}\left(cosB+cosC\right)=cosA+2\sqrt{3}cos\dfrac{B+C}{2}cos\dfrac{B-C}{2}\)
\(\le cosA+2\sqrt{3}cos\dfrac{B+C}{2}=-2sin^2\dfrac{A}{2}+2\sqrt{3}sin\dfrac{A}{2}+1\)
Xét hàm \(f\left(t\right)=-2t^2+2\sqrt{3}sint+1\) với \(t\in\left(0;1\right)\)
\(f'\left(t\right)=-4t+2\sqrt{3}=0\Rightarrow t=\dfrac{\sqrt{3}}{2}\)
\(f\left(0\right)=1\) ; \(f\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{5}{2}\) ; \(f\left(1\right)=2\sqrt{3}-1\)
\(\Rightarrow B_{max}=\dfrac{5}{2}\)
\(\Rightarrow A\ge\dfrac{1+2\sqrt{3}}{2}-\dfrac{5}{4}=\dfrac{4\sqrt{3}-3}{4}\)
a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)
b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)
c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)
\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)
d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)
\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)
e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)
\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)
\(=\left(3-1\right)^{106}=2^{106}\)
f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)
\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)
\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)
g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)
\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)
\(=2023^{4\sqrt{2}}\)
Ta chứng minh BĐT sau cho các số dương:
\(x^5+y^5\ge xy\left(x^3+y^3\right)\)
\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)
\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)
Áp dụng:
\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)
Tương tự và cộng lại:
\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)
\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)
\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)