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a ) \(2a^2+b^2+c^2\ge2a\left(b+c\right)\)
\(\Leftrightarrow a^2-2ab+b^2+a^2-2ac+c^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2\ge0\)
\(\LeftrightarrowĐPCM.\)
b ) \(a^2+2b^2+12\ge2b\left(3-a\right)\)
\(\Leftrightarrow a^2+2b^2+12\ge6b-2ab\)
\(\Leftrightarrow a^2+2ab+b^2+b^2-6b+9+3\ge0\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b-3\right)^2+3\ge0\)
\(\LeftrightarrowĐPCM.\)
c ) \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3\)
\(\Leftrightarrow a^2+2a+1+b^2+2b+1+c^2+2c+1\ge0\)
\(\Leftrightarrow\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2\ge0\)
\(\LeftrightarrowĐPCM.\)
a)theo cauchy ta có
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\a^2+c^2\ge2ac\end{matrix}\right.\)
\(\Leftrightarrow2a^2+b^2+c^2\ge2a\left(b+c\right)\Rightarrowđpcm\)
câu b) xem lại đề , tôi nghĩ phải > 0 mới đúng
c) theo cauchy ta có
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\a^2+c^2\ge2ac\\b^2+c^2\ge2bc\end{matrix}\right.\)
cộng lại, rút 2 đi suy ra đpcm
a ) \(x^2+4y^2+3z^2+14\ge2x+12y+6z\)
\(\Leftrightarrow x^2-2x+1+4y^2-12y+9+3z^2-6z+3+1\ge0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+3\left(z-1\right)^2+1\ge0\)
\(\LeftrightarrowĐPCM.\)
b ) \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+c^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\LeftrightarrowĐPCM.\)
a) \(x^2+4y^2+3z^2+14\ge2x+12y+6z\)
\(\Rightarrow x^2+4y^2+3z^2+14-2x-12y-6z\ge0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+3\left(z^2-2z+1\right)+1\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(2y-3\right)^2+3\left(z-1\right)^2\ge-1\)
Xem lại đề
b)
\(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(\Rightarrow3a^2+3b^2+3c^2\ge\left(a+b+c\right)^2\)
\(\Rightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\) *Đúng*
Dấu "=" xảy ra khi: \(a=b=c\)
c) theo bđt cauchy ta có
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+1\ge2b\\a^2+1\ge2a\end{matrix}\right.\)
cộng hết lại rút 2 đi \(\Rightarrowđpcm\)
Mẫu là abc nó lại khác nó dễ hơn thế này nhiều vì khi đó mẫu và tử sẽ hết abc
Nice proof, nhưng đã quy đồng là phải thế này :v
\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)
\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)
Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:
\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)
Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)
Áp dụng BĐT này ta có:
\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
3: =>a^2c^2+a^2d^2+b^2c^2+b^2d^2>=a^2c^2+2abcd+b^2d^2
=>a^2d^2-2abcd+b^2c^2>=0
=>(ad-bc)^2>=0(luôn đúng)
a) \(\dfrac{a^2+2}{\sqrt{a^2+1}}=\dfrac{a^2+1+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\dfrac{1}{\sqrt{a^2+1}}\ge2\)
b) Tương tự