5.1.a.
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{xy}{xy}}=2\)
Dấu "=" xảy ra khi \(x=y\)
5.1.b.
\(P=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=\frac{9}{1}=9\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
5.2.a.
\(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow3a^2-3ab+3b^2\ge a^2+ab+b^2\)
\(\Leftrightarrow2\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow2\left(a-b\right)^2\ge0\) (luôn đúng)
5.2.b
Ta có: \(\frac{a^3}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\frac{a+b}{3}=\frac{2a}{3}-\frac{b}{3}\)
Tương tự: \(\frac{b^3}{b^2+bc+c^2}\ge\frac{2b}{3}-\frac{c}{3}\) ; \(\frac{c^3}{c^2+ca+a^2}\ge\frac{2c}{3}-\frac{a}{3}\)
Cộng vế với vế:
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge\frac{a+b+c}{3}\)
Dấu "=" xảy ra khi \(a=b=c\)
5.3.a
\(a+b\ge2\sqrt{ab}\Leftrightarrow a-2\sqrt{ab}+b\ge0\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
5.3.b
\(x\left(\frac{1}{2015}+\frac{1}{x-2015}\right)\ge x\left(\frac{4}{2015+x-2015}\right)=\frac{4x}{x}=4\)
Dấu "=" xảy ra khi \(x-2015=2015\Rightarrow x=4030\)
5.4.a
\(a^5+b^5\ge a^3b^2+a^2b^3\)
\(\Leftrightarrow a^3\left(a^2-b^2\right)-b^3\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a^3-b^3\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^2+ab+b^2\right)\ge0\) (luôn đúng với a;b dương)
Dấu "=" xảy ra khi \(a=b\)
5.3.b
\(\frac{ab}{a^5+b^5+ab}\le\frac{ab}{a^3b^2+a^2b^3+ab}=\frac{1}{a^2b+ab^2+1}=\frac{c}{a^2bc+ab^2c+c}=\frac{c}{a+b+c}\)
Tương tự: \(\frac{bc}{b^5+c^5+bc}\le\frac{a}{a+b+c}\) ; \(\frac{ca}{c^5+a^5+ca}\le\frac{b}{a+b+c}\)
Cộng vế với vế: \(P\le\frac{a+b+c}{a+b+c}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
5.4.a
\(M=\frac{1}{2}\left(2x^2+2y^2-2xy-2x-2y+2\right)\)
\(M=\frac{1}{2}\left[\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\right]\ge0\)
\(M_{min}=0\) khi \(x=y=1\)
5.4.b
Đề bài sai
5.5.a
\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\frac{b^2}{c+a}+\frac{c+a}{4}\ge b\) ; \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
Cộng vế với vế:
\(VT+\frac{a+b+c}{2}\ge a+b+c\Rightarrow VT\ge\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
5.5.b
\(5x-4\sqrt{x}-2y\sqrt{x}+y^2+1=0\)
\(\Leftrightarrow x-2y\sqrt{x}+y^2+4x-4\sqrt{x}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-y\right)^2+\left(2\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-y=0\\2\sqrt{x}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=\frac{1}{2}\end{matrix}\right.\)
5.6.a
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
5.6.b
Áp dụng BĐT bên trên:
\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{2}{b}\)
Tương tự: \(\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{2}{a}\) ; \(\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{2}{c}\)
Cộng vế với vế:
\(2\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Leftrightarrow\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Dấu "=" xảy ra khi \(a=b=c\)
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