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Lời giải:
Ta có:
\(\frac{a^8+b^8+c^8}{a^3b^3c^3}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow a^8+b^8+c^8\geq a^2b^2c^2(ab+bc+ac)(*)\)
Áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} a^8+b^8\geq 2a^4b^4\\ b^8+c^8\geq 2b^4c^4\\ c^8+a^8\geq 2c^4a^4\end{matrix}\right.\Rightarrow a^8+b^8+c^8\geq a^4b^4+b^4c^4+c^4a^4\)
Tiếp tục áp dụng AM-GM:
\(a^8+b^8+a^4b^4+c^8\geq 4\sqrt[4]{a^{12}b^{12}c^8}=4a^3b^3c^2\)
\(b^8+c^8+b^4c^4+a^8\geq 4b^3c^3a^2\)
\(c^8+a^8+c^4a^4+b^8\geq 4c^3a^3b^2\)
Cộng lại: \(3(a^8+b^8+c^8)+(a^4b^4+b^4c^4+c^4a^4)\geq 4a^2b^2c^2(ab+bc+ca)\)
Mà \(a^8+b^8+c^8\geq a^4b^4+b^4c^4+c^4a^4\Rightarrow 4(a^8+b^8+c^8)\geq 4a^2b^2c^2(ab+bc+ac)\)
hay \(a^8+b^8+c^8\geq a^2b^2c^2(ab+bc+ac)\Rightarrow (*)\) đúng
Ta có đpcm.
a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)
\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)
b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)
\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)
c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)
\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{\left ( \frac{a}{bc} \right )^2}{\frac{1}{c}}+\frac{\left ( \frac{b}{ca} \right )^2}{\frac{1}{a}}+\frac{\left ( \frac{c}{ab} \right )^2}{\frac{1}{b}}\geq \frac{\left ( \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)
\(\Leftrightarrow \text{VT}\geq \frac{\left ( \frac{a^2+b^2+c^2}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)
Theo hệ quả của BĐT AM-GM thì:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{\left ( \frac{ab+bc+ac}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
Áp dụngk BĐt cô-si, ta có
\(\frac{a^2}{b^2c}+\frac{b^2}{c^2a}+\frac{1}{a}\ge3.\frac{1}{c}\)
Tương tự , rồi cộng vào, ta có
\(2A+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow A\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(ĐPCM\right)\)
^_^
1.Ta có :\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^2-xy+y^2\) (do x+y=1)
\(=\dfrac{3}{4}\left(x-y\right)^2+\dfrac{1}{4}\left(x+y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)\(=\dfrac{1}{4}.1=\dfrac{1}{4}\)
Dấu "=" xảy ra khi :\(x=y=\dfrac{1}{2}\)
Vậy \(x^3+y^3\ge\dfrac{1}{4}\)
2.
a) Sửa đề: \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì \(a,b\ge0\))
Đẳng thức xảy ra \(\Leftrightarrow a=b\)
b) Lần trước mk giải rồi nhá
3.
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel\(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)+3}=\dfrac{9}{3+3}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{y+1}=\dfrac{1}{z+1}\\x+y+z=3\end{matrix}\right.\Leftrightarrow x=y=z=1\)
b) \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{x}{2\sqrt{x^2.1}}+\dfrac{y}{2\sqrt{y^2.1}}+\dfrac{z}{2\sqrt{z^2.1}}\)
\(=\dfrac{x}{2x}+\dfrac{y}{2y}+\dfrac{z}{2z}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)
\(vì:a,b,c>0\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)
\(Cosi:\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\ge\dfrac{2}{\dfrac{a+b}{2}}=\dfrac{4}{a+b}\)
\(\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{4}{a+b}+\dfrac{4}{a+c}\right)\le\dfrac{1}{16}\left(\dfrac{8}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}.tươngtự:\dfrac{4}{a+b+2c}\le\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{2c};\dfrac{4}{a+2b+c}\le\dfrac{1}{4a}+\dfrac{1}{2b}+\dfrac{1}{2c}.\text{cộng vế theo vế ta được:}\dfrac{4}{a+2b+c}+\dfrac{4}{2a+b+c}+\dfrac{4}{a+b+2c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(\text{đpcm}\right)\)
Áp dụng BĐT \(\dfrac{1}{x+y+z+t}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)\) với các số dương
Ta có: \(\dfrac{4}{a+a+b+c}\le\dfrac{4}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\dfrac{4}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)
\(\dfrac{4}{a+2b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)
Cộng vế với vế:
\(\dfrac{4}{2a+b+c}+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\dfrac{1}{a}\ge\dfrac{2}{b}+\dfrac{8}{2a-b}\)\(\Leftrightarrow\dfrac{1}{a}-\dfrac{2}{b}-\dfrac{8}{2a-b}\ge0\)
\(\Leftrightarrow\dfrac{b\left(2a-b\right)}{ab\left(2a-b\right)}-\dfrac{2a\left(2a-b\right)}{ab\left(2a-b\right)}-\dfrac{8ab}{ab\left(2a-b\right)}\ge0\)
\(\Leftrightarrow\dfrac{b\left(2a-b\right)-2a\left(2a-b\right)-8ab}{ab\left(2a-b\right)}\ge0\)
\(\Leftrightarrow\dfrac{2ab-b^2+2ab-4a^2-8ab}{ab\left(2a-b\right)}\ge0\)
\(\Leftrightarrow\dfrac{-4a^2-4ab-b^2}{ab\left(2a-b\right)}\ge0\)\(\Leftrightarrow\dfrac{4a^2+4ab+b^2}{ab\left(b-2a\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(2a+b\right)^2}{ab\left(b-2a\right)}\ge0\forall a>0;b< 0\)