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pennypc123456789   1
N 5 minutes ago by 1475393141xj
Given $x,y,z$ be the positive real number. Prove that

$\frac{2xy}{\sqrt{2xy(x^2+y^2)}} + \frac{2yz}{\sqrt{2yz(y^2+z^2)}} + \frac{2xz}{\sqrt{2xz(x^2+z^2)}} \le \frac{2(x^2+y^2+z^2) + xy+yz+xz}{x^2+y^2+z^2}$
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pennypc123456789
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Interesting inequalities
sqing   6
N Apr 16, 2025 by sqing
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
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sqing
Apr 16, 2025
sqing
Apr 16, 2025
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Interesting inequalities
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sqing
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sqing
#1 h  Apr 16, 2025, 3:36 AM
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 6:13 AM
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sqing
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#2 h  Apr 16, 2025, 3:37 AM
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc =4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4(k-1) $$Where $ \frac{16}{9}\geq k\geq 0. $
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 6:16 AM
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#3 h  Apr 16, 2025, 4:02 AM
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Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc =4$ . Prove that
$$k(a+b+c) +a^2-bc\geq 4(k-1) $$Where $ 2.6647417 \geq k\geq 0. $
$$ k(a+b+c) +a^2-ab - bc\geq 4(k-1) $$Where $ 1.968711\geq k\geq 0. $
This post has been edited 2 times. Last edited by sqing, Apr 16, 2025, 6:18 AM
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lbh_qys
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#4 h  Apr 16, 2025, 5:47 AM
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sqing wrote:
Let $ a,b,c\geq 0 $ and $ a^2+b^2+c^2+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$

When $a=b=\sqrt{2}, c=0$, then $\frac{16}{9}(a+b+c) -ab-bc= \frac{32\sqrt{2}}{9} - 2 < \frac{28}{9}$.
This post has been edited 1 time. Last edited by lbh_qys, Apr 16, 2025, 5:47 AM
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#5 h  Apr 16, 2025, 6:14 AM
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Sorry .Thank lbh_qys.
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
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lbh_qys
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lbh_qys
#6 h  Apr 16, 2025, 10:13 AM
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sqing wrote:
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc=4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4\sqrt{k(k+1)}-(k+4)$$Where $ k\geq \frac{16}{9}. $
$$ \frac{16}{9}(a+b+c) -ab-bc\geq \frac{28}{9}$$
sqing wrote:
Let $ a,b,c\geq 0 $ and $ ab+bc+ca+abc =4$ . Prove that
$$k(a+b+c) -ab-bc\geq 4(k-1) $$Where $ \frac{16}{9}\geq k\geq 0. $


If \(b > k\), since \(ab+bc \leq ab+bc+ca+abc = 4\), we have \( a+c \le \frac{4}{b}\), therefore

\[ k(a+b+c) - ab - bc + (k+4) = kb + (a+c)(k-b) + k + 4 \ge kb + \frac{4}{b}(k-b) + k + 4 \]\[ = kb + \frac{4k}{b} + k \ge 2\sqrt{kb \cdot \frac{4k}{b}} + k = 5k \]
If \(b \le k\), since
\[ 0 = b(a+c) + (b+1)ac - 4 \le b(a+c) + \frac{(b+1)(a+c)^2}{4} - 4 = (a+c+4)\bigl(a+c + ab + bc - 4\bigr), \]
we have \( a+c \geq \frac{4}{1+b} \), thus
\[ k(a+b+c) - ab - bc + (k+4) = k(b+1) + (a+c)(k-b) + 4 \ge k(b+1) + \frac{4}{b+1}(k-b) + 4 \]\[ = k(b+1) + \frac{4(k+1)}{b+1} \ge 2\sqrt{ k(b+1) \cdot \frac{4(k+1)}{b+1}} = 4 \sqrt{k(k+1)} \]
Hence we have

\[ k(a+b+c) -ab-bc \geq \min\left\{5k, 4\sqrt{k(k+1)} \right\} -(k+4) \]
This post has been edited 8 times. Last edited by lbh_qys, Apr 16, 2025, 10:57 AM
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#7 h  Apr 16, 2025, 12:29 PM
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Very very nice.Thank lbh_qys.
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