cho a+b = 1. c/m: a^4+b^4 \(\ge\) 1/8
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Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
áp dụng bất đằng thức buinhia
\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Leftrightarrow1\le2\left(a^2+b^2\right)\Rightarrow a^2+b^2\ge\frac{1}{2}\)
\(\left(a^2+b^2\right)^2\le\left(\left(a^2\right)^2+\left(b^2\right)^2\right)2\Leftrightarrow\left(\frac{1}{2}\right)^2\le2\left(a^4+b^4\right)\Rightarrow a^4+b^4\ge\frac{1}{8}\)
bài cuối tương tự
a, \(a^2+b^2\ge\frac{1}{2}\)
Với mọi a, b ta có:
\(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+2ab+b^2\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
Mà a + b = 1 \(\Rightarrow2\left(a^2+b^2\right)\ge1\)
\(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\)
Vậy \(a^2+b^2\ge\frac{1}{2}\)( đpcm )
Các câu b, c tương tự
Áp dụng bđt Cauchy-Schwarz:
\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
\(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{1}{2}\right)^2}{2}=\dfrac{1}{8}\)
\(a^8+b^8\ge\dfrac{\left(a^4+b^4\right)^2}{2}\ge\dfrac{\left(\dfrac{1}{8}\right)^2}{2}=\dfrac{1}{128}\)
Câu a : \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow\left(a-b\right)^2\ge0\)
a: =>2a^2+2b^2>=a^2+2ab+b^2
=>a^2-2ab+b^2>=0
=>(a-b)^2>=0(luôn đúng)
c: =>3a^2+3b^2+3c^2>=a^2+b^2+c^2+2ab+2bc+2ac
=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0
=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)
a/
\(VT\ge\frac{\frac{1}{2}\left(a+b\right)^2}{a+b}+\frac{\frac{1}{2}\left(b+c\right)^2}{b+c}+\frac{\frac{1}{2}\left(c+a\right)^2}{c+a}=a+b+c\ge3\sqrt[3]{abc}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ Ta có: \(x^4+y^4\ge\frac{1}{2}\left(x^2+y^2\right)\left(y^2+y^2\right)\ge xy\left(x^2+y^2\right)\)
\(\Rightarrow VT\le\frac{1}{a+bc\left(b^2+c^2\right)}+\frac{1}{b+ca\left(a^2+c^2\right)}+\frac{1}{c+ab\left(a^2+b^2\right)}\)
\(VT\le\frac{1}{a+\frac{1}{a}\left(b^2+c^2\right)}+\frac{1}{b+\frac{1}{b}\left(a^2+c^2\right)}+\frac{1}{c+\frac{1}{c}\left(a^2+b^2\right)}\)
\(VT\le\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}=\frac{a+b+c}{a^2+b^2+c^2}\)
\(VT\le\frac{a+b+c}{\frac{1}{3}\left(a+b+c\right)^2}=\frac{3}{a+b+c}\le\frac{3}{3\sqrt[3]{abc}}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(P=\dfrac{\sqrt{a-1}}{a}+\dfrac{\sqrt{b-4}}{b}+\dfrac{\sqrt{c-9}}{c}=\dfrac{1.\sqrt{a-1}}{a}+\dfrac{2.\sqrt{b-4}}{2b}+\dfrac{3.\sqrt{c-9}}{3c}\)
Áp dụng hằng đẳng thức \(xy\le\dfrac{x^2+y^2}{2}\) ta được
\(P\le\dfrac{1+a-1}{2a}+\dfrac{4+b-4}{4b}+\dfrac{9+c-9}{6c}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{11}{12}\)
\(\Rightarrow P_{max}=\dfrac{11}{12}\) khi \(\left\{{}\begin{matrix}\sqrt{a-1}=1\\\sqrt{b-4}=2\\\sqrt{c-9}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=8\\c=18\end{matrix}\right.\)
dễ Cm được x² +y² ≥ (x+y)²/2
<=> x² +y² ≥ 1/2(x² +y²) + xy
<=> 1/2(x² +y²) -xy ≥ 0
<=> 1/2(x-y)² ≥ 0 ( luôn đúng )
vậy x² + y² ≥ (x+y)²/2 = 1/2
tương tự thì
x^4 + y^4 ≥ (x² +y²)²/2 ≥ (1/2)²/2 = 1/8
vậy x^4 + y^4 ≥ 1/8
dấu = xảy ra <=> x=y=1/2
Đế sai : a = 0 ; b = 1 => a + b = 0 +1 = 1 nhưng
a^4 + b^4 = 0^4 + 1^4 = 0 + 1 = 1 khác 1/8