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1.
C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)
\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)
\(\Leftrightarrow3a^3+3a^2b-3ab^2-3b^3\ge0\)
\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)
\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)
Theo bài ra: \(a^3-b^3\ge3a-3b-4\)
\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)
Thật vậy:
\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)
\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)
\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))
2.
\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)
\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)
\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)
\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)
d/ \(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) (luôn đúng)
e/ \(\Leftrightarrow a^6+b^6+a^5b+ab^5\ge a^6+b^5+a^4b^2+a^2b^4\)
\(\Leftrightarrow a^5b-a^4b^2+ab^5-a^2b^4\ge0\)
\(\Leftrightarrow a^4b\left(a-b\right)-ab^4\left(a-b\right)\ge0\)
\(\Leftrightarrow ab\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow ab\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) (luôn đúng)
f/ \(\frac{a^6}{b^2}+a^2b^2\ge2\sqrt{\frac{a^8b^2}{b^2}}=2a^4\) ; \(\frac{b^6}{a^2}+a^2b^2\ge2b^4\)
\(\Rightarrow\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge2a^4+2b^4-2a^2b^2\)
\(\Leftrightarrow\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge a^4+b^4+\left(a^4+b^4-2a^2b^2\right)\)
\(\Leftrightarrow\frac{a^6}{b^2}+\frac{b^6}{a^2}\ge a^4+b^4+\left(a^2-b^2\right)^2\ge a^4+b^4\)
a/ \(VT=a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\)
\(VT=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2\)
\(VT\ge6\sqrt[6]{a^6b^6c^6}=6\left|abc\right|\ge6abc\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ \(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4e^2\ge4ab+4ac+4ad+4ae\)
\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2e\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(\frac{a}{2}=b=c=d=e\)
c/ \(\Leftrightarrow\frac{a^3+b^3}{2}\ge\frac{a^3+b^3+3a^2b+3ab^2}{8}\)
\(\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b\)
1.
\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)
\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)
Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá
2.
\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
Đặt \(x+y+z=t\Rightarrow0< t\le1\)
\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
3.
\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)
Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)
Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)
4.
ĐKXĐ: \(-2\le x\le2\)
\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)
\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)
\(y_{min}=-2\) khi \(x=-2\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
\(\frac{a^3}{\left(1-a\right)^2}+\frac{1-a}{8}+\frac{1-a}{8}\ge3\sqrt[3]{\frac{a^3}{\left(1-a\right)^2}.\frac{\left(1-a\right)}{8}.\frac{1-a}{8}}=\frac{3a}{4}\)
Suy ra \(\frac{a^3}{1-a^2}\ge\frac{3a}{4}-\frac{\left(1-a\right)}{4}=\frac{4a-1}{4}\)
Tương tự hai BĐT còn lại rồi cộng theo vế:
\(A\ge\frac{4\left(a+b+c\right)-3}{4}=\frac{1}{4}\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Câu 2)
Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)
\(\Leftrightarrow9\ge4\left(ab+2\right)\)
\(\Rightarrow9\ge4ab+8\)
\(\Rightarrow1\ge4ab\)
Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow a^2+2ab+b^2\ge4ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )
Câu 3)
Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
Mà \(a+b+c=1\)
\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)
\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng bất đẳng thức Cô-si
\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)
\(\Rightarrow\) ĐPCM
1) Áp dụng BĐT AM-GM: \(VT\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9=VP\)
Đẳng thức xảy ra khi $a=b=c.$
2) Từ (1) suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{3^2}{a+b+c}+\frac{1^2}{d}\ge\frac{\left(3+1\right)^2}{a+b+c+d}=VP\)
Đẳng thức..
3) Ta có \(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\) với $a,b,c>0.$
Cho $c=1$ ta nhận được bất đẳng thức cần chứng minh.
4) Đặt \(a=x^2,b=y^2,S=x+y,P=xy\left(S^2\ge4P\right)\) thì cần chứng minh $$(x+y)^8 \geqq 64x^2 y^2 (x^2+y^2)^2$$
Hay là \(S^8\ge64P^2\left(S^2-2P\right)^2\)
Tương đương với $$(-4 P + S^2)^2 ( 8 P S^2 + S^4-16 P^2 ) \geqq 0$$
Đây là điều hiển nhiên.
5) \(3a^3+\frac{7}{2}b^3+\frac{7}{2}b^3\ge3\sqrt[3]{3a^3.\left(\frac{7}{2}b^3\right)^2}=3\sqrt[3]{\frac{147}{4}}ab^2>9ab^2=VP\)
6) \(VT=\sqrt[4]{\left(\sqrt{a}+\sqrt{b}\right)^8}\ge\sqrt[4]{64ab\left(a+b\right)^2}=2\sqrt{2\left(a+b\right)\sqrt{ab}}=VP\)
Có thế thôi mà nhỉ:v